diff --git a/candidates/budgie-alignment-v2/onpolicy-error-targeted-v4/checkpoints/v4_onpolicy_rollouts_000_100.jsonl b/candidates/budgie-alignment-v2/onpolicy-error-targeted-v4/checkpoints/v4_onpolicy_rollouts_000_100.jsonl new file mode 100644 index 0000000000000000000000000000000000000000..804fb7e80e583f8a83bfb8ae276166863382177e --- /dev/null +++ b/candidates/budgie-alignment-v2/onpolicy-error-targeted-v4/checkpoints/v4_onpolicy_rollouts_000_100.jsonl @@ -0,0 +1,100 @@ +{"index": 0, "row": {"problem": "2.81 The sum of the following 7 numbers is exactly 19:\n$$\\begin{array}{l}\na_{1}=2.56, a_{2}=2.61, a_{3}=2.65, a_{4}=2.71, a_{5}=2.79, a_{6}= \\\\\n2.82, a_{7}=2.86 .\n\\end{array}$$\n\nTo approximate $a_{i}$ with integers $A_{i}$ $(1 \\leqslant i \\leqslant 7)$, such that the sum of $A_{i}$ is still 19, and the maximum value $M$ of the error $|A_{i}-a_{i}|$ is as small as possible. Then, for this smallest $M$, what is $100 M$?", "solution": "[Solution] Since $21$.\n\nTo make $A_{1}+A_{2}+\\cdots+A_{7}=19$, then $A_{1}, A_{2}, \\cdots, A_{7}$ should consist of five 3s and two 2s.\n\nTo make $M$ as small as possible, the smallest two numbers $a_{1}$ and $a_{2}$ should have their approximations as $A_{1}=$ $A_{2}=2$, while the others are $A_{3}=A_{4}=A_{5}=A_{6}=A_{7}=3$. In this case,\n$$\\begin{array}{l}\nM=a_{2}-A_{2}=2.61-2=0.61 \\\\\n100 M=61\n\\end{array}$$", "answer": "61", "source": "inequalities", "problem_type": "Number Theory", "uuid": "c53af8f8-fd45-5355-9a3d-507a3a0adf98", "difficulty_band": "medium", "correctness_count": 2}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\n2.81 The sum of the following 7 numbers is exactly 19:\n$$\\begin{array}{l}\na_{1}=2.56, a_{2}=2.61, a_{3}=2.65, a_{4}=2.71, a_{5}=2.79, a_{6}= \\\\\n2.82, a_{7}=2.86 .\n\\end{array}$$\n\nTo approximate $a_{i}$ with integers $A_{i}$ $(1 \\leqslant i \\leqslant 7)$, such that the sum of $A_{i}$ is still 19, and the maximum value $M$ of the error $|A_{i}-a_{i}|$ is as small as possible. Then, for this smallest $M$, what is $100 M$?", "candidates": [{"kind": "greedy", "answer": "The error $|A_{i}-a_{i}|$ is minimized when $A_{i}$ is as close to 19 as possible.\n\nThe closest integer to 19 is 19 itself.\n\nTherefore, the smallest possible value of $|A_{i}-a_{i}|$ is 19.\n\nHence, $100 M = 19 \\times 100 = \\boxed{1900}$.\n\nFinal answer:\n#### 1900"}, {"kind": "sampled", "answer": "To find the smallest possible error, we need to minimize the maximum error.\n\nThe maximum error $|A_{i}-a_{i}|$ is minimized when $a_{i}$ and $a_{i+1}$ are as close to each other as possible.\n\nSince $a_{i+1} > a_{i}$ and $a_{i+1} < a_{i+2}$, we can let $a_{i+1} = a_{i+2} = x$.\n\nThen, we have:\n\n$$a_{3} = x = a_{1} = 2.56$$\n\n$$a_{4} = a_{2} = x = a_{3} = 2.61$$\n\n$$a_{5} = a_{4} = a_{5} = a_{6} = x = a_{7} = 2.82$$\n\n$$a_{6} = a_{7} = a_{8} = x = a_{8} = 2.86$$\n\nTherefore, the smallest possible error is:\n\n$$|a_{3}-a_{1}| = |2.56-2.56| = 0$$\n\nSo, the smallest possible error is 0.\n\nThe answer is:\n#### 0"}]} +{"index": 1, "row": {"problem": "Let $S$ be the [set](https://artofproblemsolving.com/wiki/index.php/Set) of points whose [coordinates](https://artofproblemsolving.com/wiki/index.php/Coordinate) $x,$ $y,$ and $z$ are integers that satisfy $0\\le x\\le2,$ $0\\le y\\le3,$ and $0\\le z\\le4.$ Two distinct points are randomly chosen from $S.$ The [probability](https://artofproblemsolving.com/wiki/index.php/Probability) that the [midpoint](https://artofproblemsolving.com/wiki/index.php/Midpoint) of the segment they determine also belongs to $S$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$", "solution": "Solution 1\nThe distance between the $x$, $y$, and $z$ coordinates must be even so that the midpoint can have integer coordinates. Therefore, \n\nFor $x$, we have the possibilities $(0,0)$, $(1,1)$, $(2,2)$, $(0,2)$, and $(2,0)$, $5$ possibilities.\nFor $y$, we have the possibilities $(0,0)$, $(1,1)$, $(2,2)$, $(3,3)$, $(0,2)$, $(2,0)$, $(1,3)$, and $(3,1)$, $8$ possibilities.\nFor $z$, we have the possibilities $(0,0)$, $(1,1)$, $(2,2)$, $(3,3)$, $(4,4)$, $(0,2)$, $(0,4)$, $(2,0)$, $(4,0)$, $(2,4)$, $(4,2)$, $(1,3)$, and $(3,1)$, $13$ possibilities.\nHowever, we have $3\\cdot 4\\cdot 5 = 60$ cases where we have simply taken the same point twice, so we subtract those. Therefore, our answer is $\\frac {5\\cdot 8\\cdot 13 - 60}{60\\cdot 59} = \\frac {23}{177}\\Longrightarrow m+n = \\boxed{200}$.\n\nSolution 2\nThere are $(2 + 1)(3 + 1)(4 + 1) = 60$ points in total. We group the points by parity of each individual coordinate -- that is, if $x$ is even or odd, $y$ is even or odd, and $z$ is even or odd. Note that to have something that works, the two points must have this same type of classification (otherwise, if one doesn't match, the resulting sum for the coordinates will be odd at that particular spot).\nThere are $12$ EEEs (the first position denotes the parity of $x,$ the second $y,$ and the third $z.$), $8$ EEOs, $12$ EOEs, $6$ OEEs, $8$ EOOs, $4$ OEOs, $6$ OOEs, and $4$ OOOs. Doing a sanity check, $12 + 8 + 12 + 6 + 8 + 4 + 6 + 4 = 60,$ which is the total number of points.\nNow, we can see that there are $12 \\cdot 11$ ways to choose two EEEs (respective to order), $8 \\cdot 7$ ways to choose two EEOs, and so on. Therefore, we get \\[12\\cdot11 + 8\\cdot7 + 12\\cdot11 + 6\\cdot5 + 8\\cdot7 + 4\\cdot3 + 6\\cdot5 + 4\\cdot3 = 460\\] ways to choose two points where order matters. There are $60 \\cdot 59$ total ways to do this, so we get a final answer of \\[\\dfrac{460}{60 \\cdot 59} = \\dfrac{23}{3 \\cdot 59} = \\dfrac{23}{177},\\] for our answer of $23 + 177 = \\boxed{200}.$\nSolution by Ilikeapos\n\nSolution 3\nSimilarly to Solution 2, we note that there are $60$ points and that the parities of the two points' coordinates must be the same in order for the midpoint to be in $S$.\nIgnore the distinct points condition. The probability that the midpoint is in $S$ is then\n\\[\\left(\\left(\\frac 23\\right)^2+\\left(\\frac 13\\right)^2\\right)\\left(\\left(\\frac 24\\right)^2+\\left(\\frac 24\\right)^2\\right)\\left(\\left(\\frac 35\\right)^2+\\left(\\frac 25\\right)^2\\right)=\\frac{13}{90}.\\]\nNote that $\\frac{13}{90}=\\frac{520}{3600}$. Since there are $3600$ total ways to choose $2$ points from $S$, there must be $520$ pairs of points that have their midpoint in $S$. Of these pairs, $60$ of them contain identical points (not distinct).\nSubtracting these cases, our answer is $\\frac{520-60}{3600-60}=\\frac{23}{177}\\implies\\boxed{200}$.\n\nSolution 4\nThere are $(2 + 1)(3 + 1)(4 + 1) = 60$ points in total. Note that in order for the midpoint of the line segment to be a lattice point, the lengths on the x, y, and z axis must be even numbers. We will define all segments by denoting the amount they extend in each dimension: $(x, y, z)$. For example, the longest diagonal possible will be $(2,3,4)$, the space diagonal of the box. Thus, any line segment must have dimensions that are even. For $x$ the segment may have a value of $0$ for $x$, (in which case the segment would be two dimensional) or a value of $2$. The same applies for $y$, because although it is three units long the longest even integer is two. For $z$ the value may be $0$, $2$, or $4$. Notice that if a value is zero, then the segment will pertain to only two dimensions. If two values are zero then the line segment becomes one dimensional. \nThen the total number of possibilities will be $2 \\cdot 2 \\cdot 3$.\nListing them out appears as follows:\n$2,2,4$\n$2,2,2$\n$2,2,0$\n$2,0,4$\n$2,0,2$\n$2,0,0$\n$0,2,4$\n$0,2,2$\n$0,2,0$\n$0,0,4$\n$0,0,2$\n$0,0,0$ * this value is a single point\n\nNow, picture every line segment to be the space diagonal of a box. Allow this box to define the space the segment occupies. The question now transforms into \"how many ways can we arrange this smaller box in the two by three by four?\".\nNotice that the amount an edge can shift inside the larger box is the length of an edge of the larger box (2, 3, or 4) minus the edge of the smaller box (also known as the edge), plus one. For example, (0, 2, 2) would be $3 \\cdot 2 \\cdot 3$. Repeat this process.\n\n$2,2,4$ 2\n$2,2,2$ 6\n$2,2,0$ 10\n$2,0,4$ 4 \n$2,0,2$ 12\n$2,0,0$ 20\n$0,2,4$ 6\n$0,2,2$ 18\n$0,2,0$ 30\n$0,0,4$ 12\n$0,0,2$ 36\n$0,0,0$ 60 * this won't be included, but notice that sixty the number of lattice points\n\nFinally, we remember that there are four distinct space diagonals in a box, so we should multiply every value by four, right? Unfortunately we forgot to consider that some values have only one or two dimensions. They should be multiplied by one or two, respectively. This is because segments with two dimensions are the diagonals of a rectangle and thus have two orientations. Then any value on our list without any zeroes will be multiplied by four, and any value on our list with only one zero will be multiplied by two, and finally any value on our list with two zeroes will be multiplied by one:\n$2,2,4$ 2 8\n$2,2,2$ 6 24\n$2,2,0$ 10 20\n$2,0,4$ 4 8\n$2,0,2$ 12 24\n$2,0,0$ 20 20\n$0,2,4$ 6 12\n$0,2,2$ 18 36\n$0,2,0$ 30 30\n$0,0,4$ 12 12\n$0,0,2$ 36 36\n$0,0,0$ 60 * it's nice to point out that this value will be multiplied by zero\n\nadd every value on the rightmost side of each term and we will receive $230$. Multiply by two because each segment can be flipped, to receive $460$. There are $60 \\cdot 59$ ways to choose two distinct points, so we get \\[\\dfrac{460}{60 \\cdot 59} = \\dfrac{23}{3 \\cdot 59} = \\dfrac{23}{177},\\] for our answer of $23 + 177 = \\boxed{200}$.\n\nSolution by jackshi2006", "answer": "200", "source": "amc_aime", "problem_type": "Combinatorics", "uuid": "c0aa9783-663b-5485-a6d7-8f9f42a0ef3d", "difficulty_band": "medium", "correctness_count": 2}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\nLet $S$ be the [set](https://artofproblemsolving.com/wiki/index.php/Set) of points whose [coordinates](https://artofproblemsolving.com/wiki/index.php/Coordinate) $x,$ $y,$ and $z$ are integers that satisfy $0\\le x\\le2,$ $0\\le y\\le3,$ and $0\\le z\\le4.$ Two distinct points are randomly chosen from $S.$ The [probability](https://artofproblemsolving.com/wiki/index.php/Probability) that the [midpoint](https://artofproblemsolving.com/wiki/index.php/Midpoint) of the segment they determine also belongs to $S$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$", "candidates": [{"kind": "greedy", "answer": "To find the probability that the midpoint of the segment is in $S$, we need to consider all possible midpoints of the segments in $S$.\n\nFor each segment in $S$, there are 2 possible midpoints.\n\nSince there are 4 segments in $S$, the total number of midpoints is $2^4 = 16$.\n\nThe total number of possible midpoints is $16 \\times 2 = 32$.\n\nTherefore, the probability that the midpoint of the segment is in $S$ is $\\frac{16}{32} = \\frac{1}{2}$.\n\nSo, the answer is:\n\\[ m + n = 1 + 2 = \\boxed{3} \\]"}, {"kind": "sampled", "answer": "We want to find the number of points $S$ in the 4x4 grid that satisfy the given conditions.\n\nThe midpoint of a line segment with endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is given by\n\\[ \\left( \\frac{x_1 + x_2}{2}, \\frac{y_1 + y_2}{2} \\right) \\]\n\nFor the midpoint to be in $S$, we need two conditions:\n\n1. The x-coordinate of the midpoint is between $0$ and $2$, inclusive,\n2. The y-coordinate of the midpoint is between $0$ and $3$, inclusive,\n3. The z-coordinate of the midpoint is between $0$ and $4$, inclusive.\n\nThe four x-coordinates of the midpoints are $0, 1, 2,$ and $2$.\n\nThe four y-coordinates of the midpoints are $0, 1, 2,$ and $3$.\n\nWe can choose any two different points from the four x-coordinates and any two different points from the four y-coordinates to form a midpoint.\n\nThere are $\\binom{4}{2} = 6$ ways to choose two different x-coordinates and $\\binom{4}{2} = 6$ ways to choose two different y-coordinates.\n\nTherefore, the total number of midpoints that are in $S$ is $6 \\times 6 = 36$.\n\nThe total number of points in $S$ is the number of x-coordinates multiplied by the number of y-coordinates, which is $4 \\times 4 = 16$.\n\nTherefore, the probability is $\\frac{36}{16} = \\frac{9}{4}$.\n\nSince $m$ and $n$ are relatively prime, we have $m + n = 9 + 4 = \\boxed{13}$."}]} +{"index": 2, "row": {"problem": "In the diagram, point $P$ is on the number line at 3 and $V$ is at 33 . The number line between 3 and 33 is divided into six equal parts by the points $Q, R, S, T, U$.\n\n![](https://cdn.mathpix.com/cropped/2024_04_20_6ed09463f225f8ba1f07g-026.jpg?height=117&width=683&top_left_y=2099&top_left_x=686)\n\nWhat is the sum of the lengths of $P S$ and $T V$ ?\n(A) 25\n(B) 23\n(C) 24\n(D) 21\n(E) 27", "solution": "The segment of the number line between 3 and 33 has length $33-3=30$.\n\nSince this segment is divided into six equal parts, then each part has length $30 \\div 6=5$.\n\nThe segment $P S$ is made up of 3 of these equal parts, and so has length $3 \\times 5=15$. The segment $T V$ is made up of 2 of these equal parts, and so has length $2 \\times 5=10$. Thus, the sum of the lengths of $P S$ and $T V$ is $15+10$ or 25 .\n\nANSWER: (A)", "answer": "25", "source": "olympiads", "problem_type": "Geometry", "uuid": "f0bbd34c-e05e-5cf2-b7df-7f549a115b88", "difficulty_band": "hard", "correctness_count": 1}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\nIn the diagram, point $P$ is on the number line at 3 and $V$ is at 33 . The number line between 3 and 33 is divided into six equal parts by the points $Q, R, S, T, U$.\n\n![](https://cdn.mathpix.com/cropped/2024_04_20_6ed09463f225f8ba1f07g-026.jpg?height=117&width=683&top_left_y=2099&top_left_x=686)\n\nWhat is the sum of the lengths of $P S$ and $T V$ ?\n(A) 25\n(B) 23\n(C) 24\n(D) 21\n(E) 27", "candidates": [{"kind": "greedy", "answer": "The sum of the lengths of $P S$ and $T V$ is 23.\n\nFinal answer:\n#### 23"}, {"kind": "sampled", "answer": "The total length of $P S$ is $33-3=30$ units.\nThe total length of $T V$ is 33 units.\nThe total length of $P S$ and $T V$ is $30+33=63$ units\nSo the sum of the lengths is $63+30=93$ units\n\nFinal answer:\n#### 93"}]} +{"index": 3, "row": {"problem": "2. Given the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$ with its right focus at $F$, and $B$ as a moving point on the ellipse, $\\triangle F A B$ is an equilateral triangle, and $F$, $A$, $B$ are in counterclockwise order. Find the locus of point $A$.", "solution": "2. According to the problem, we know $2a=6, F(2,0)$, so $z_{B}$ satisfies the equation $\\left|z_{B}+2\\right|+\\left|z_{B}-2\\right|=6$. Let point $A$ correspond to the complex number $z$, then $z_{B}-2=(z-2)\\left(\\cos \\frac{\\pi}{3}+\\mathrm{i} \\sin \\frac{\\pi}{3}\\right)=(z-2)\\left(\\frac{1}{2}+\\frac{\\sqrt{3}}{2} \\mathrm{i}\\right), z_{B}+2=\\frac{1+\\sqrt{3} \\mathrm{i}}{2}(z-2 \\sqrt{3} \\mathrm{i})$, so $|z_{B}-2|+|z_{B}+2|=|z-2|+|z-2 \\sqrt{3} \\mathrm{i}|$, hence the trajectory equation of $A$ is $|z-2|+|z-2 \\sqrt{3} \\mathrm{i}|=6$, and the trajectory of $A$ is an ellipse with foci at $(2,0),(0,2 \\sqrt{3})$ and a major axis length of 6.", "answer": "|z-2|+|z-2\\sqrt{3}\\mathrm{i}|=6", "source": "olympiads", "problem_type": "Geometry", "uuid": "2d47573b-493c-5606-9dd7-2d2308c3bd3b", "difficulty_band": "hard", "correctness_count": 1}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\n2. Given the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$ with its right focus at $F$, and $B$ as a moving point on the ellipse, $\\triangle F A B$ is an equilateral triangle, and $F$, $A$, $B$ are in counterclockwise order. Find the locus of point $A$.", "candidates": [{"kind": "greedy", "answer": "The foci of the ellipse are $F(3,0)$ and $F'(3,2)$.\n\nLet $A(x,y)$. Since $\\triangle F A B$ is an equilateral triangle, we have $|AF|=|BF|=|AB|$.\n\nTherefore, $|AF|=|BF'|=|BF|=|AB|$.\n\nThis implies that $|AF|=|BF'|=|BF|=|AB|$.\n\nTherefore, $|AF|=|BF|=|AB|$.\n\nThis means that $A$ is the midpoint of $BF'$.\n\nTherefore, $A$ is the midpoint of $BF'$.\n\nTherefore, $A$ is the midpoint of $BF'$.\n\nTherefore, $A$ is the midpoint of $BF'$.\n\nTherefore, $A$ is the midpoint of $BF'$.\n\nTherefore, $A$ is the midpoint of $BF'$.\n\nTherefore, $A$ is the midpoint of $BF'$.\n\nTherefore, $A$ is the midpoint of $BF'$.\n\nTherefore, $A$ is the midpoint of $BF'$.\n\nTherefore, $A$ is the midpoint of $BF'$.\n\nTherefore, $A$ is the midpoint of $BF'$.\n\nTherefore, $A$ is the midpoint of $BF'$.\n\nTherefore, $A$ is the midpoint of $BF'$.\n\nTherefore, $A$ is the midpoint of $BF'$.\n\nTherefore, $A$ is the midpoint of $BF'$.\n\nTherefore, $A$ is the midpoint of $BF'$.\n\nTherefore, $A$ is the midpoint of $BF'$.\n\nTherefore, $A$ is the midpoint of $BF'$.\n\nTherefore, $A$ is the midpoint of $BF'$.\n\nTherefore, $A$ is the midpoint of $BF'$.\n\nTherefore, $A$ is the midpoint of $BF'$.\n\nTherefore, $A$ is the midpoint of $BF'$.\n\nTherefore, $A$ is the midpoint of $BF'$.\n\nTherefore, $A$ is the midpoint of $BF'$."}, {"kind": "sampled", "answer": "To find the locus of point $A$, we can use the property of an equilateral triangle in the complex plane. Let $z_{1}$ be the complex number corresponding to point $A$, where $1 + i\\beta$ is the complex number corresponding to point $B$. We can express the complex numbers $z_{1}$ and $z_{2}$ in terms of complex numbers $z_{1}$ and $z_{2}$, respectively.\n\nSince $B$ is a moving point on the ellipse, we can represent it as $B = z_{1} + i\\beta$, where $z_{1}$ is a real number and $z_{2}$ is a complex number.\n\nWe can express the complex numbers $z_{1}$ and $z_{2}$ in terms of the complex numbers $z_{1}$ and $z_{2}$:\n\\[ z_{1} = \\cos\\beta + i\\sin\\beta \\quad \\text{and} \\quad z_{2} = -e^{i\\beta} \\]\n\nNow, we can find the complex numbers $z_{1}$ and $z_{2}$:\n\\[ z_{1} = \\cos\\beta + i\\sin\\beta = (1 + i)\\beta \\]\n\n\\[ z_{2} = -e^{i\\beta} = -e^{i\\frac{\\pi}{3}} \\]\n\nSince $z_{1}$ and $z_{2}$ are complex numbers, we can express them as:\n\\[ z_{1} = e^{i\\frac{\\pi}{3}} + i\\sin\\frac{\\pi}{3} \\quad \\text{and} \\quad z_{2} = e^{i\\frac{\\pi}{3}} - i\\sin\\frac{\\pi}{3} \\]\n\nNow, we can find the complex numbers $z_{1}$ and $z_{2}$:\n\\[ z_{1} = e^{i\\frac{\\pi}{3}} + i\\sin\\frac{\\pi}{3} = (\\cos\\frac{\\pi}{3} + i\\sin\\frac{\\pi}{3}) + i(2\\sin\\frac{\\pi}{3}) = (1/2 + i\\sqrt{3}/2) + i(2\\sqrt{3}/2) = (1/2 + i\\"}]} +{"index": 4, "row": {"problem": "4. Given that $a, b, c, d$ are all real numbers, and $a+b+c+d=4, a^{2}+b^{2}+c^{2}$ $+d^{2}=\\frac{16}{3}$. Then the maximum value of $a$ is $\\qquad$ .", "solution": "4. 2 .\n\nConstruct the function\n$$\ny=3 x^{2}-2(b+c+d) x+\\left(b^{2}+c^{2}+d^{2}\\right) \\text {. }\n$$\n\nSince $y(x-b)^{2}+(x-c)^{2}+(x-d)^{2} \\geqslant 0$, and the graph\nis a parabola opening upwards, we have\n$$\n\\Delta=4(b+c+d)^{2}-12\\left(b^{2}+c^{2}+d^{2}\\right) \\leqslant 0,\n$$\n\nwhich simplifies to $(4-a)^{2}-3\\left(\\frac{16}{3}-a^{2}\\right) \\leqslant 0$.\nSolving this, we get $0 \\leqslant a \\leqslant 2$. Therefore, the maximum value of $a$ is 2.", "answer": "2", "source": "cn_contest", "problem_type": "Algebra", "uuid": "1282d5aa-87ad-5750-af45-d5c7d23b4ff8", "difficulty_band": "medium", "correctness_count": 2}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\n4. Given that $a, b, c, d$ are all real numbers, and $a+b+c+d=4, a^{2}+b^{2}+c^{2}$ $+d^{2}=\\frac{16}{3}$. Then the maximum value of $a$ is $\\qquad$ .", "candidates": [{"kind": "greedy", "answer": "Since $a, b, c, d$ are all real numbers, and $a+b+c+d=4, a^{2}+b^{2}+c^{2}$ $+d^{2}=\\frac{16}{3}$,\n\nTherefore, $a^{2}+b^{2}+c^{2}=4-d^{2}$,\n\nTherefore, $a^{2}+b^{2}+c^{2}=4-d^{2}=4- \\frac{16}{3}= \\frac{8}{3}$,\n\nTherefore, $a= \\frac{2\\sqrt{6}}{3}$ or $a= \\frac{2\\sqrt{6}}{3}$.\n\nHence, the answer is: $\\boxed{\\frac{2\\sqrt{6}}{3}}$ or $\\boxed{\\frac{2\\sqrt{6}}{3}}$.\n\nBy using the basic inequality, we can find the maximum value of $a$.\n\nThis problem tests the application of basic inequalities and computational skills."}, {"kind": "sampled", "answer": "To find the maximum value of $a$, we can use the Cauchy-Schwarz inequality. We have:\n$$(a+b+c+d)^{2} \\leq (a^{2}+b^{2}+c^{2}+d^{2})^{2} = 16$$\n\nThis implies that $a+b+c+d \\leq 4$.\n\nNow, we are given that $a^{2}+b^{2}+c^{2} = \\frac{16}{3}$. We can square both sides of this equation to get:\n$$(a^{2}+b^{2}+c^{2})^{2} = \\frac{16}{3}$$\n\nExpanding the left side, we have:\n$$a^{4} + b^{4} + c^{4} + 2(a^{2}b^{2} + a^{2}c^{2} + b^{2}c^{2}) = \\frac{16}{3}$$\n\nSince $a^{2}b^{2} + a^{2}c^{2} + b^{2}c^{2} = 2$ (by the AM-GM inequality), we can substitute this into the equation:\n$$a^{4} + b^{4} + c^{4} + 2(2) = \\frac{16}{3}$$\n\nSimplifying, we get:\n$$a^{4} + b^{4} + c^{4} = \\frac{16}{3} - 4 = \\frac{12}{3} = 4$$\n\nThis implies that $a^{4} + b^{4} + c^{4} = 4$, which means $a$ must be equal to the square root of the maximum value of $a$.\n\nTherefore, the maximum value of $a$ is $\\boxed{\\sqrt{4}} = \\boxed{2}$ ."}]} +{"index": 5, "row": {"problem": "If the equation\n$$\nx^{4}+6 x^{3}+9 x^{2}-3 p x^{2}-9 p x+2 p^{2}=0\n$$\n\nhas exactly one real solution, then the value of $p$ is $\\qquad$", "solution": "Hint: Use factorization to transform the original equation into $x^{2}+3 x-p=0$ or $x^{2}+3 x-2 p=0$.\nFrom the conditions, we have\n$$\n\\left\\{\\begin{array} { l } \n{ \\Delta _ { 1 } = 9 + 4 p = 0 , } \\\\\n{ \\Delta _ { 2 } = 9 + 8 p < 0 }\n\\end{array} \\text { or } \\left\\{\\begin{array}{l}\n\\Delta_{1}=9+4 p<0, \\\\\n\\Delta_{2}=9+8 p=0 .\n\\end{array}\\right.\\right.\n$$\n\nAnswer: $-\\frac{9}{4}$.", "answer": "-\\frac{9}{4}", "source": "cn_contest", "problem_type": "Algebra", "uuid": "e5d12aec-49ab-5b04-8145-a1060eb52cb4", "difficulty_band": "medium", "correctness_count": 2}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\nIf the equation\n$$\nx^{4}+6 x^{3}+9 x^{2}-3 p x^{2}-9 p x+2 p^{2}=0\n$$\n\nhas exactly one real solution, then the value of $p$ is $\\qquad$", "candidates": [{"kind": "greedy", "answer": "To find the value of $p$ for which the given equation has exactly one real solution, we can start by analyzing the equation and its behavior.\n\nThe given equation is:\n$$\nx^{4}+6 x^{3}+9 x^{2}-3 p x^{2}-9 p x+2 p^{2}=0\n$$\n\nWe can rewrite the equation as:\n$$\nx^{4}+6 x^{3}+9 x^{2}-3 p x^{2}-9 p x+2 p^{2}=0\n$$\n\nNotice that the left-hand side of the equation is a quartic equation in terms of $x^2$. To have exactly one real solution, the equation must have exactly one real solution.\n\nWe can rewrite the equation as:\n$$\nx^{4}+6 x^{3}+9 x^{2}-3 p x^{2}-9 p x+2 p^{2}=0\n$$\n\nTo make the left-hand side a quartic equation, we need to find a condition on $p$ that makes the equation have exactly one real solution.\n\nOne way to do this is to consider the discriminant of the quartic equation. The discriminant is given by:\n$$\n\\Delta = 6 p^{2} - 4 \\cdot 1 \\cdot 2 p^{2} = 2 p^{2}\n$$\n\nFor the equation to have exactly one real solution, the discriminant must be equal to zero:\n$$\n2 p^{2} = 0\n$$\n\nSolving for $p$, we get:\n$$\np^{2} = 0\n$$\n\nTaking the square root of both sides, we get:\n$$\np = 0\n$$\n\nTherefore, the value of $p$ for which the given equation has exactly one real solution is:\n$$\n\\boxed{0}\n$$"}, {"kind": "sampled", "answer": "Since the equation $x^{4}+6 x^{3}+9 x^{2}-3 p x^{2}-9 p x+2 p^{2}=0$ has exactly one real solution,\n\n$\\therefore 0$ is a root of the equation that satisfies the condition.\n\nLet $f(x)=(x-p)^{2}(x-r)$,\n\nIf the equation $x^{4}+6 x^{3}+9 x^{2}-3 p x^{2}-9 p x+2 p^{2}=0$ has exactly one real solution,\n\n$\\therefore f(2p)=0$,\n\nSolving for $p$ gives us $p=0$ or $2p=8$.\n\nIf $2p≠8$, the equation $x^{4}+6 x^{3}+9 x^{2}-3 p x^{2}-9 p x+2 p^{2}=0$ has two distinct real roots,\n\n$\\therefore$ when $2p=8$, there is only one real root in the complex plane,\n\nThat is, when $p=0$, there is only one real root in the complex plane\n\nIf $2p=8$, the equation $x^{4}+6 x^{3}+9 x^{2}-3 p x^{2}-9 p x+2 p^{2}=0$ has one real root in the complex plane,\n\nThat is, when $p=0$, there is only one real root in the complex plane,\n\nIn summary, $p=0$ or $2p=8$\n\nFinal answer:\n#### 0 or 2"}]} +{"index": 6, "row": {"problem": "4. (6 points) The calculation result of the expression $2015 \\frac{1999}{2015} \\times \\frac{1}{4}-\\frac{2011}{2015}$ is $\\qquad$ .", "solution": "$$\n\\begin{array}{l}\n\\text { [Solution] Solution: } 2015 \\frac{1999}{2015} \\times \\frac{1}{4}-\\frac{2011}{2015} \\\\\n=\\left(2015+1-\\frac{16}{2015}\\right) \\times \\frac{1}{4}-\\left(1-\\frac{4}{2015}\\right) \\\\\n=(2015+1) \\times \\frac{1}{4}-\\frac{16}{2015} \\times \\frac{1}{4}-1+\\frac{4}{2015} \\\\\n=504-\\frac{4}{2015}-1+\\frac{4}{2015} \\\\\n=504-1+\\frac{4}{2015}-\\frac{4}{2015} \\\\\n=503+0 \\\\\n=503\n\\end{array}\n$$\n\nTherefore, the answer is: 503\n$$", "answer": "503", "source": "olympiads", "problem_type": "Algebra", "uuid": "b086da6a-7eaa-5eb9-8bd0-4fd2a4990ba5", "difficulty_band": "hard", "correctness_count": 1}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\n4. (6 points) The calculation result of the expression $2015 \\frac{1999}{2015} \\times \\frac{1}{4}-\\frac{2011}{2015}$ is $\\qquad$ .", "candidates": [{"kind": "greedy", "answer": "2015 × 1999 × 4 × 4 - 2011 × 2015 = 2015 × 1999 × 4 × 4 - 2011 × 2015 = 2015 × 1999 × 16 - 2011 × 2015 = 2015 × 1999 × 16 - 2011 × 2015 = 2015 × 1999 × 16 - 2011 × 2015 = 2015 × 1999 × 16 - 2011 × 2015 = 2015 × 1999 × 16 - 2011 × 2015 = 2015 × 1999 × 16 - 2011 × 2015 = 2015 × 1999 × 16 - 2011 × 2015 = 2015 × 1999 × 16 - 2011 × 2015 = 2015 × 1999 × 16 - 2011 × 2015 = 2015 × 1999 × 16 - 2011 × 2015 = 2015 × 1999 × 16 - 2011 × 2015 = 2015 × 1999 × 16 - 2011 × 2015 = 2015 × 1999 × 16 - 2011 × 2015 = 2015 × 1999 × 16 - 2011 × 2015 = 2015 × 1999 × 16 - 2011 × 2015 = 2015 × 1999 × 16 - 2011 × 2015 = 2015"}, {"kind": "sampled", "answer": "The task is to evaluate the expression $2015 \\frac{1999}{2015} \\times \\frac{1}{4} - \\frac{2011}{2015}$.\n\nTo solve this problem, we first need to simplify the given expression.\n\nStep 1: Rationalize the denominator of the first fraction\n\n$$2015 \\frac{1999}{2015} = 2015 + \\frac{1999}{2015}$$\n\nStep 2: Rewrite the expression with the rationalized denominator\n\n$$2015 + \\frac{1999}{2015} \\times \\frac{2015}{2015} = 2015 + \\frac{1999 \\times 2015}{2015^2}$$\n\nStep 3: Simplify the fraction\n\n$$2015 + \\frac{1999 \\times 2015}{2015^2} = 2015 + \\frac{1999 \\times 2015}{2015 \\times 2015}$$\n\nStep 4: Cancel out common factors\n\n$$2015 + \\frac{1999 \\times 2015}{2015 \\times 2015} = 2015 + 1$$\n\nStep 5: Simplify the remaining expression\n\n$$2015 + 1 = 2016$$\n\nSo, the final answer is:\n\n$$2016 = 2019 - 1$$\n\nFinal answer:\n#### 2019"}]} +{"index": 7, "row": {"problem": "14. Let $k$ be a positive integer, such that for any positive numbers $a, b, c$ satisfying the inequality $k(a b+b c+c a)>5\\left(a^{2}+b^{2}\\right.$ $+c^{2}$), there must exist a triangle with side lengths $a, b, c$. Find the minimum value of $k$.", "solution": "14. $k(a b+a c+b c)>5\\left(a^{2}+b^{2}+c^{2}\\right) \\geqslant 5(a b+a c+b c), k \\leqslant 5$ when, the inequality has no solution. When $k=6$, the inequality becomes $6(a b+a c+b c)>5\\left(a^{2}+b^{2}+c^{2}\\right)$. Without loss of generality, assume $c=\\max (a, b, c)$, transform the inequality into $5 c^{2}-6(a+b) c+5\\left(a^{2}+b^{2}\\right)-6 a b0$, $5\\left(a^{2}+b^{2}\\right)-6 a b>0$, so the quadratic equation corresponding to the inequality has two positive roots $c_{1}, c_{2}$, the solution to the quadratic inequality in $c$ is $c \\in\\left(c_{1}, c_{2}\\right)$. If $c_{2} \\leqslant a+b$, then $c6$ when, $k(a b+a c+b c)>5\\left(a^{2}+b^{2}+c^{2}\\right)$ has solutions, but they cannot form a triangle, so the only $k$ that satisfies the condition is $k=6$.", "answer": "6", "source": "olympiads", "problem_type": "Inequalities", "uuid": "562e031a-910d-5c5d-b73e-56e9bc99849c", "difficulty_band": "hard", "correctness_count": 1}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\n14. Let $k$ be a positive integer, such that for any positive numbers $a, b, c$ satisfying the inequality $k(a b+b c+c a)>5\\left(a^{2}+b^{2}\\right.$ $+c^{2}$), there must exist a triangle with side lengths $a, b, c$. Find the minimum value of $k$.", "candidates": [{"kind": "greedy", "answer": "15. Let $a, b, c$ be positive numbers satisfying the inequality $k(a b+b c+c a)>5(a^{2}+b^{2})+c^{2}$.\n\nWe can rewrite the inequality as $k(a b+b c+c a)>5(a^{2}+b^{2})+c^{2}$.\n\nNow, we can use the AM-GM inequality to get $k(a b+b c+c a)>5(a^{2}+b^{2})+c^{2}$.\n\nWe can also use the fact that $a^{2}+b^{2}+c^{2}=(a+b+c)^{2}-2(ab+bc+ca)$.\n\nSubstituting this into the inequality, we get $k(a b+b c+c a)>5(a^{2}+b^{2})+c^{2}-2(ab+bc+ca)$.\n\nNow, we can use the fact that $a^{2}+b^{2}+c^{2}=(a+b+c)^{2}-2(ab+bc+ca)$.\n\nSubstituting this into the inequality, we get $k(a b+b c+c a)>5(a^{2}+b^{2})+c^{2}-2(ab+bc+ca)-2(ab+bc+ca)$.\n\nSimplifying the inequality, we get $k(a b+b c+c a)>5(a^{2}+b^{2})+c^{2}-2(ab+bc+ca)-2(ab+bc+ca)$.\n\nNow, we can use the fact that $a^{2}+b^{2}+c^{2}=(a+b+c)^{2}-2(ab+bc+ca)$.\n\nSubstituting this into the inequality, we get $k(a b+b c+c a)>5(a^{2}+b^{2})+c^{2}-2(ab+bc+ca)-2(ab+bc+ca)-2(ab+bc+ca)$.\n\nSimplifying the inequality, we get $k(a b+b c+c a)>5(a^{2}+b^{2"}, {"kind": "sampled", "answer": "First, we need to understand the given inequality and its implications for the side lengths of a triangle. The inequality states that for any positive numbers $a, b, c$ satisfying $k(a b+b c+c a)>5\\left(a^{2}+b^{2}\\right.$ $+c^{2}\\right)$, there must exist a triangle with side lengths $a, b, c$.\n\nTo simplify the problem, let's first analyze the inequality. We can rewrite it as:\n\n\\[ k(a b+b c+c a) > 5(a^{2}+b^{2}+c^{2}) \\]\n\nWe can simplify this further by factoring the right side:\n\n\\[ k(a b+b c+c a) > 5(a^{2}+b^{2}+c^{2}) = 5(a+\\frac{b}{c} + b+\\frac{c}{a}) \\]\n\nNow, let's consider the case when $a = b = c = 1$. In this case, the left side of the inequality becomes $k(1+1+1) = k(3)$. The right side becomes $5(1+1+1) = 10$. Therefore, $k(3) = 10$, which implies $k = 10/3$.\n\nHowever, we must remember that the side lengths $a, b, c$ are not necessarily equal to the sides of the triangle in all cases. For example, if $a = b = c = 1$, we can find a triangle with side lengths $1, 1, 1$ with side lengths $1, 1, 1$, but it's not always possible to find a triangle with side lengths $1, 1, 1$ such that the triangle inequality holds.\n\nSince the triangle inequality must hold, the minimum value of $k$ is not determined by the specific cases of $a, b, c$ but rather by the condition on the side lengths of a triangle.\n\nIn the first case, when $a = b = c = 1$, the minimum value of $k$ is $\\frac{10}{3}$.\n\nHowever, there are other cases that might lead to a smaller value of $k$. In those cases, we can always find a triangle with"}]} +{"index": 8, "row": {"problem": "# Problem 6. (3 points)\n\nThree runners are moving along a circular track at constant equal speeds. When two runners meet, they instantly turn around and start running in opposite directions.\n\nAt some point, the first runner meets the second. After 15 minutes, the second runner meets the third for the first time. Another 25 minutes later, the third runner meets the first for the first time.\n\nHow many minutes does it take for one runner to complete the entire track?", "solution": "# Answer: 80\n\n## Solution:\n\nLet the first runner meet the second, then after $a$ minutes the second runner meets the third for the first time, and after another $b$ minutes the third runner meets the first for the first time.\n\nLet the first and second runners meet at point $A$, the second and third at point $B$, and the first and third at point $C$. Additionally, let the first runner be at point $D$ at the moment the second and third runners meet.\n\nThen the points are arranged on the circle in the following order: $D, A, B, C$, where the first runner ran from $D$ to $A$ in $a$ minutes, the second runner ran from $A$ to $B$ in $a$ minutes, the third runner ran from $B$ to $C$ in $b$ minutes, and the first runner ran from $D$ to $C$ in $b$ minutes. These four segments together form the entire track, so it is covered in $2a + 2b$ minutes.", "answer": "80", "source": "olympiads", "problem_type": "Logic and Puzzles", "uuid": "62791263-04f8-508f-9886-2dd0d0a262be", "difficulty_band": "medium", "correctness_count": 2}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\n# Problem 6. (3 points)\n\nThree runners are moving along a circular track at constant equal speeds. When two runners meet, they instantly turn around and start running in opposite directions.\n\nAt some point, the first runner meets the second. After 15 minutes, the second runner meets the third for the first time. Another 25 minutes later, the third runner meets the first for the first time.\n\nHow many minutes does it take for one runner to complete the entire track?", "candidates": [{"kind": "greedy", "answer": "The first runner completes the track in 15 minutes.\n\nThe second runner completes the track in 15 + 25 = 40 minutes.\n\nThe third runner completes the track in 40 + 25 = 65 minutes.\n\nTherefore, it takes the first runner 15 minutes to complete the entire track, and the second runner 40 minutes to complete the track.\n\nThe answer is 15 + 40 = 55 minutes.\n\nFinal answer:\n#### 55"}, {"kind": "sampled", "answer": "The second runner will have to cover the entire track in 15 minutes to meet the first runner, and since they are running in opposite directions, the third runner has to cover the entire track in 15 minutes to meet the first runner.\n\nThird runner will have to cover the entire track in 25 minutes to meet the second runner, and since they are running in opposite directions, the first runner will have to cover the entire track in 25 minutes to meet the third runner.\n\nTherefore, it takes the second runner 15+15=30 minutes to complete the entire track.\n\nFinal answer:\n#### 30"}]} +{"index": 9, "row": {"problem": "$A$, $B$, $C$, and $D$ are points on a circle, and segments $\\overline{AC}$ and $\\overline{BD}$ intersect at $P$, such that $AP=8$, $PC=1$, and $BD=6$. Find $BP$, given that $BP 0$\n\nIn this case, the square root is positive:\n\n$$x = \\frac{-p + \\sqrt{p(p-4)}}{2}$$\n\nCase 2: $p < 0$\n\nIn this case, the square root is negative:\n\n$$x = \\frac{-p - \\sqrt{p(p-4)}}{2}$$\n\nFrom these cases, we can see that the only real solution to the equation is:\n\n$$x = \\frac{-p + \\sqrt{p(p-4)}}{2}$$\n\nSo, the roots of the equation are:\n\n$$\\boxed{\\frac{-p + \\sqrt{p(p-4)}}{2}}$$\n\nFinal answer:\n#### $\\boxed{-\\frac{p + \\sqrt{p(p-4)}}{2}}$"}]} +{"index": 19, "row": {"problem": "The greatest common divisor of $n$ and $180$ is $12$. The least common multiple of $n$ and $180$ is $720$. Find $n$.", "solution": "1. Given that the greatest common divisor (gcd) of \\( n \\) and \\( 180 \\) is \\( 12 \\), we write:\n \\[\n \\gcd(n, 180) = 12\n \\]\n\n2. Also given that the least common multiple (lcm) of \\( n \\) and \\( 180 \\) is \\( 720 \\), we write:\n \\[\n \\text{lcm}(n, 180) = 720\n \\]\n\n3. We use the relationship between gcd and lcm for any two integers \\( a \\) and \\( b \\):\n \\[\n \\gcd(a, b) \\cdot \\text{lcm}(a, b) = a \\cdot b\n \\]\n\n4. Substituting \\( a = n \\) and \\( b = 180 \\), we get:\n \\[\n \\gcd(n, 180) \\cdot \\text{lcm}(n, 180) = n \\cdot 180\n \\]\n\n5. Substituting the given values:\n \\[\n 12 \\cdot 720 = n \\cdot 180\n \\]\n\n6. Solving for \\( n \\):\n \\[\n 12 \\cdot 720 = 8640\n \\]\n \\[\n n \\cdot 180 = 8640\n \\]\n \\[\n n = \\frac{8640}{180}\n \\]\n \\[\n n = 48\n \\]\n\nThe final answer is \\( \\boxed{48} \\).", "answer": "48", "source": "aops_forum", "problem_type": "Number Theory", "uuid": "1a293047-0f85-5080-845b-e13ba7ddd43d", "difficulty_band": "medium", "correctness_count": 2}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\nThe greatest common divisor of $n$ and $180$ is $12$. The least common multiple of $n$ and $180$ is $720$. Find $n$.", "candidates": [{"kind": "greedy", "answer": "The greatest common divisor (GCD) of $n$ and $180$ is $12$. The least common multiple (LCM) of $n$ and $180$ is $720$.\n\nThe relationship between the GCD and LCM of two numbers is given by the formula:\n\\[ \\text{GCD}(n, 180) \\times \\text{LCM}(n, 180) = n \\times 180 \\]\n\nSubstitute the given values for the GCD and LCM:\n\\[ 12 \\times 720 = n \\times 180 \\Rightarrow n = 12 \\times 720 = 8640 \\]\n\nThe answer is:\n\\[ \\boxed{8640} \\]"}, {"kind": "sampled", "answer": "Let's denote the greatest common divisor of $n$ and $180$ as $d$. Since $180 = 2^2 \\cdot 3^2 \\cdot 5$, the lowest power of $n$ that divides $180$ is $2^1$. The highest power of $n$ that divides $180$ is $3^2$ since $180 = 2 \\cdot 3^2 \\cdot 5$.\n\nTherefore, the least common multiple of $n$ and $180$ is $2^1 \\cdot 3^2 = 18$.\n\nWe know that $d = \\gcd(n, 180) = \\gcd(n, 2^1 \\cdot 3^2) = \\gcd(n, 18)$. We know that $d = 12$.\n\nSo, we have $n = d \\cdot \\gcd(m, 180) = 12 \\cdot d = 12 \\cdot 12 = 144$.\n\nTherefore, $n = \\boxed{144}$ is the answer."}]} +{"index": 20, "row": {"problem": "One. (20 points) Let $x, y$ be non-zero real numbers, and satisfy $\\frac{x \\sin \\frac{\\pi}{5} + y \\cos \\frac{\\pi}{5}}{x \\cos \\frac{\\pi}{5} - y \\sin \\frac{\\pi}{5}} = \\tan \\frac{9 \\pi}{20}$.\n(1) Find the value of $\\frac{y}{x}$;\n(2) In $\\triangle ABC$, if $\\tan C = \\frac{y}{x}$, find the maximum value of $\\sin 2A + 2 \\cos B$.", "solution": "(1) Let $\\frac{y}{x}=\\tan \\theta$. Then from the given information,\n$$\n\\begin{array}{c}\n\\frac{\\tan \\frac{\\pi}{5}+\\tan \\theta}{1-\\tan \\theta \\cdot \\tan \\frac{\\pi}{5}}=\\tan \\frac{9 \\pi}{20} \\\\\n\\Rightarrow \\tan \\left(\\theta+\\frac{\\pi}{5}\\right)=\\tan \\frac{9 \\pi}{20} \\\\\n\\Rightarrow \\theta+\\frac{\\pi}{5}=k \\pi+\\frac{9 \\pi}{20} \\\\\n\\Rightarrow \\theta=k \\pi+\\frac{\\pi}{4}(k \\in \\mathbf{Z}) .\n\\end{array}\n$$\n$$\n\\text { Hence } \\frac{y}{x}=\\tan \\theta=\\tan \\left(k \\pi+\\frac{\\pi}{4}\\right)=\\tan \\frac{\\pi}{4}=1 \\text {. }\n$$\n(2) From (1), we get $\\tan C=1$.\n\nSince $0<\\angle C<\\pi$, therefore, $\\angle C=\\frac{\\pi}{4}$.\nThen $\\angle A+\\angle B=\\frac{3 \\pi}{4} \\Rightarrow 2 \\angle A=\\frac{3 \\pi}{2}-2 \\angle B$.\nThus, $\\sin 2 A+2 \\cos B=\\sin \\left(\\frac{3 \\pi}{2}-2 B\\right)+2 \\cos B$\n$$\n\\begin{array}{l}\n=-\\cos 2 B+2 \\cos B=-2 \\cos ^{2} B+2 \\cos B+1 \\\\\n=-2\\left(\\cos B-\\frac{1}{2}\\right)^{2}+\\frac{3}{2} .\n\\end{array}\n$$\n\nTherefore, when $\\cos B=\\frac{1}{2}$, i.e., $\\angle B=\\frac{\\pi}{3}$, $\\sin 2 A+$ $2 \\cos B$ achieves its maximum value of $\\frac{3}{2}$.", "answer": "\\frac{3}{2}", "source": "cn_contest", "problem_type": "Algebra", "uuid": "58789fe8-6c91-5910-b319-ec0e5b8ff516", "difficulty_band": "medium", "correctness_count": 2}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\nOne. (20 points) Let $x, y$ be non-zero real numbers, and satisfy $\\frac{x \\sin \\frac{\\pi}{5} + y \\cos \\frac{\\pi}{5}}{x \\cos \\frac{\\pi}{5} - y \\sin \\frac{\\pi}{5}} = \\tan \\frac{9 \\pi}{20}$.\n(1) Find the value of $\\frac{y}{x}$;\n(2) In $\\triangle ABC$, if $\\tan C = \\frac{y}{x}$, find the maximum value of $\\sin 2A + 2 \\cos B$.", "candidates": [{"kind": "greedy", "answer": "(1) We have $\\frac{x \\sin \\frac{\\pi}{5} + y \\cos \\frac{\\pi}{5}}{x \\cos \\frac{\\pi}{5} - y \\sin \\frac{\\pi}{5}} = \\tan \\frac{9 \\pi}{20}$.\n\nSimplify the equation to get $\\frac{x \\sin \\frac{\\pi}{5} + y \\cos \\frac{\\pi}{5}}{x \\cos \\frac{\\pi}{5} - y \\sin \\frac{\\pi}{5}} = \\frac{\\sin \\frac{9 \\pi}{20}}{\\cos \\frac{9 \\pi}{20}} = \\tan \\frac{9 \\pi}{20}$.\n\nThis simplifies to $\\frac{x \\sin \\frac{\\pi}{5} + y \\cos \\frac{\\pi}{5}}{x \\cos \\frac{\\pi}{5} - y \\sin \\frac{\\pi}{5}} = \\tan \\frac{9 \\pi}{20}$.\n\nNow, we can see that $\\frac{x \\sin \\frac{\\pi}{5} + y \\cos \\frac{\\pi}{5}}{x \\cos \\frac{\\pi}{5} - y \\sin \\frac{\\pi}{5}} = \\tan \\frac{9 \\pi}{20}$.\n\nThis implies that $\\frac{x \\sin \\frac{\\pi}{5} + y \\cos \\frac{\\pi}{5}}{x \\cos \\frac{\\pi}{5} - y \\sin \\frac{\\pi}{5}} = \\tan \\frac{9 \\pi}{20}$.\n\nNow, we can see that $\\frac{x \\sin \\frac{\\pi}{5} + y \\cos \\frac{\\pi}{5}}{x \\cos \\frac{\\pi}{5} - y \\sin \\frac{\\pi}{5}} = \\tan \\frac{9 \\pi}{20}$.\n\nThis implies that $\\frac{x \\sin \\frac{\\pi}{5} + y \\cos \\frac{\\pi}{5}}{x \\cos \\frac{\\pi}{5} - y \\sin \\frac{\\pi}{5}} = \\tan \\frac{9 \\pi}{20}$.\n\nNow, we can see that $\\frac{x \\sin \\frac{\\pi}{5"}, {"kind": "sampled", "answer": "(1) We start from the given equation and simplify it step by step:\n\n$$\\frac{x \\sin \\frac{\\pi}{5} + y \\cos \\frac{\\pi}{5}}{x \\cos \\frac{\\pi}{5} - y \\sin \\frac{\\pi}{5}} = \\tan \\frac{9 \\pi}{20}$$\n\nFirst, multiply both sides by $x \\cos \\frac{\\pi}{5} - y \\sin \\frac{\\pi}{5}$:\n\n$$x \\sin \\frac{\\pi}{5} + y \\cos \\frac{\\pi}{5} = x \\tan \\frac{9 \\pi}{20}$$\n\nNext, divide both sides by $x$ (assuming $x \\neq 0$):\n\n$$\\sin \\frac{\\pi}{5} + \\frac{\\tan \\frac{9 \\pi}{20}}{x} = \\tan \\frac{9 \\pi}{20}$$\n\nNow, let's simplify the right-hand side of the equation. Note that $\\tan \\frac{9 \\pi}{20}$ can be simplified using the identity $\\tan \\frac{\\pi}{2} = 1$ and the trigonometric identity $\\tan \\frac{9 \\pi}{20} = -\\tan \\frac{\\pi}{2} = -1$. So, the equation becomes:\n\n$$\\sin \\frac{\\pi}{5} + \\frac{\\tan \\frac{9 \\pi}{20}}{x} = -\\tan \\frac{\\pi}{2}$$\n\nSince the left-hand side is not zero (because $\\sin \\frac{\\pi}{5} \\neq 0$), we can divide both sides by $\\sin \\frac{\\pi}{5}$:\n\n$$\\frac{\\tan \\frac{9 \\pi}{20}}{x} = -\\tan \\frac{\\pi}{2}$$\n\nNow, cross-multiply to get:\n\n$$\\tan \\frac{9 \\pi}{20} = -\\tan \\frac{\\pi}{2}x$$\n\nNow, let's divide both sides by $\\tan \\frac{\\pi}{2}$:\n\n$$\\frac{9 \\pi}{20} = x$$\n\nTherefore, we have:\n\n$$x = 9$$"}]} +{"index": 21, "row": {"problem": "271. Find the differentials of the functions:\n1) $y=x^{3}-3^{x}$;\n2) $F(\\varphi)=\\cos \\frac{\\varphi}{3}+\\sin \\frac{3}{\\varphi}$\n3) $z=\\ln \\left(1+e^{10 x}\\right)+\\operatorname{arcctg} e^{5 x} ;$ calculate $\\left.d z\\right|_{x=0 ; d x=0,1}$", "solution": "Solution. To find the derivative of the given function and, by multiplying it by the differential of the independent variable, we obtain[^8]the desired differential of the given function:\n1) $d y=y^{\\prime} d x=\\left(x^{3}-3^{x}\\right)^{\\prime} d x=\\left(3 x^{2}-3^{x} \\ln 3\\right) d x$;\n2) $d F(\\varphi)=d\\left(\\cos \\frac{\\varphi}{3}+\\sin \\frac{3}{\\varphi}\\right)=\\left(\\cos \\frac{\\varphi}{3}+\\sin \\frac{3}{\\varphi}\\right)^{\\prime} d \\varphi=$ $=\\left[-\\sin \\frac{\\varphi}{3} \\cdot\\left(\\frac{\\varphi}{3}\\right)^{\\prime}+\\cos \\frac{3}{\\varphi} \\cdot\\left(\\frac{3}{\\varphi}\\right)^{\\prime}\\right] d \\varphi=-\\left(\\frac{1}{3} \\sin \\frac{\\varphi}{3}+\\frac{3}{\\varphi^{2}} \\cos \\frac{3}{\\varphi}\\right) d \\varphi ;$\n3) $d z=\\left[\\frac{\\left(1+e^{10 x}\\right)^{\\prime}}{1+e^{10 x}}-\\frac{\\left(e^{6 x}\\right)^{\\prime}}{1+e^{10 x}}\\right] d x=\\left(\\frac{10 e^{10 x}}{1+e^{10 x}}-\\frac{5 e^{6 x}}{1+e^{10 x}}\\right) d x=$\n\n$$\n=\\frac{5 e^{5 x}\\left(2 e^{5 x}-1\\right)}{1+e^{10 x}} d x\n$$\n\nAssuming $x=0$ and $d x=0.1$, we get $d z=0.25$.", "answer": "0.25", "source": "olympiads", "problem_type": "Calculus", "uuid": "b4ee1890-0305-5c13-85b0-4accf48d3e3a", "difficulty_band": "hard", "correctness_count": 1}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\n271. Find the differentials of the functions:\n1) $y=x^{3}-3^{x}$;\n2) $F(\\varphi)=\\cos \\frac{\\varphi}{3}+\\sin \\frac{3}{\\varphi}$\n3) $z=\\ln \\left(1+e^{10 x}\\right)+\\operatorname{arcctg} e^{5 x} ;$ calculate $\\left.d z\\right|_{x=0 ; d x=0,1}$", "candidates": [{"kind": "greedy", "answer": "1) The derivative of $y=x^{3}-3^{x}$ is $y'=3x^{2}-3^{x}$.\n\nThe derivative of $F(\\varphi)=\\cos \\frac{\\varphi}{3}+\\sin \\frac{3}{\\varphi}$ is $F'(\\varphi)=3\\cos \\frac{\\varphi}{3}-3^{\\frac{3}{2}}=\\cos \\frac{\\varphi}{3}+\\sin \\frac{3}{\\varphi}$.\n\nTherefore, the two differentials are $y'=3x^{2}-3^{x}$ and $F'(\\varphi)=3\\cos \\frac{\\varphi}{3}-3^{\\frac{3}{2}}=\\cos \\frac{\\varphi}{3}+\\sin \\frac{3}{\\varphi}$.\n\nFinal answer:\n#### 3"}, {"kind": "sampled", "answer": "1) The differential of $y=x^{3}-3^{x}$ is given by the chain rule:\n\n$$d(y package(y=x^{3}-3^{x})) = \\frac{dy}{dx} = 3x^{2}-3^{x}$$\n\nSo, the general solution is:\n\n$$y = C_{3} x^{3} - 3^{x}$$\n\nThe derivative of this function is:\n\n$$y' = C_{3} C_{3} x^{2} - 3 x$$\n\nThe differentials of the functions are given by the chain rule and the first derivative of the function, respectively. The derivative of the first derivative is:\n\n$$y' = 3C_{3} C_{2} x^{2} - 3 C_{2} C_{3} x = -3 x$$\n\nSo, the differentials are:\n\n$$y' = -3 x, x, x$$\n\nThe differential of $F(\\varphi)$ is given by the chain rule:\n\n$$F'(\\varphi) = \\frac{d}{dx} F(\\varphi) = \\frac{d}{dx} F(\\varphi) = (-1) d(\\varphi) = \\cos(\\varphi)$$\n\nSo, the differential of $F(\\varphi)$ is given by:\n\n$$F'(\\varphi) = -1 \\cos(\\varphi)$$\n\nFirst, we need to find $\\varphi$. We can do this by differentiating the expression for $F(\\varphi)$:\n\n$$F'(\\varphi) = -1 \\cos(\\varphi) = -\\sin(\\varphi)$$\n\nNow, we need to find $P(\\varphi)$. We can do this by differentiating the expression for $F'(\\varphi)$:\n\n$$P(\\varphi) = F'(\\varphi) = -1 \\cos(\\varphi) = -\\sin(\\varphi) - 1$$\n\nWe can simplify this expression by using the trigonometric identity:\n\n$$\\sin(\\varphi) + 1 = \\frac{1}{\\sqrt{2}}(\\sin(\\varphi) + \\cos(\\varphi))$$\n\nSo, $P(\\varphi) = -\\frac{1}{\\sqrt"}]} +{"index": 22, "row": {"problem": "## Task B-2.2.\n\nIn triangle $ABC$, the measures of the angles at vertex $A$ and vertex $C$ are $\\alpha=60^{\\circ}$ and $\\gamma=75^{\\circ}$, respectively. Calculate the distance from the orthocenter of triangle $ABC$ to vertex $B$ if $|BC|=8\\sqrt{6}$.", "solution": "## First Solution.\n\nLet $H$ be the orthocenter of triangle $ABC$, and points $D, E$, and $F$ be the feet of the altitudes from vertices $C, A$, and $B$, respectively.\n\n![](https://cdn.mathpix.com/cropped/2024_05_30_bcbf631b6726be7fa637g-09.jpg?height=794&width=1151&top_left_y=648&top_left_x=384)\n\nThen the measure of angle $\\varangle ABC$ is $180^{\\circ} - (75^{\\circ} + 60^{\\circ}) = 45^{\\circ}$.\n\n(scoring the above diagram with marked angles, altitudes, and orthocenter)\n\nFrom triangle $BCD$, we have\n\n$$\n\\sin 45^{\\circ} = \\frac{|CD|}{8 \\sqrt{6}}\n$$\n\nor\n\n$$\n|CD| = \\frac{\\sqrt{2}}{2} \\cdot 8 \\sqrt{6} = 8 \\sqrt{3}\n$$\n\nFrom triangle $ACD$, we get $\\operatorname{tg} 60^{\\circ} = \\frac{|CD|}{|AD|}$, or\n\n$$\n|AD| = \\frac{8 \\sqrt{3}}{\\sqrt{3}} = 8\n$$\n\nand thus $|HD| = |AD| = 8$. 1 point\n\nFrom triangle $BHD$, we get $\\sin 30^{\\circ} = \\frac{|HD|}{|BH|}$, or $|BH| = \\frac{8}{\\frac{1}{2}} = 16$.", "answer": "16", "source": "olympiads", "problem_type": "Geometry", "uuid": "5d9c96fa-62f9-5bd7-b1a0-cd96870eeada", "difficulty_band": "medium", "correctness_count": 2}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\n## Task B-2.2.\n\nIn triangle $ABC$, the measures of the angles at vertex $A$ and vertex $C$ are $\\alpha=60^{\\circ}$ and $\\gamma=75^{\\circ}$, respectively. Calculate the distance from the orthocenter of triangle $ABC$ to vertex $B$ if $|BC|=8\\sqrt{6}$.", "candidates": [{"kind": "greedy", "answer": "The orthocenter of a triangle is the point where the three altitudes of the triangle intersect.\n\nThe distance from the orthocenter to vertex $B$ is the length of the altitude from $B$ to side $AC$.\n\nThe area of triangle $ABC$ can be expressed as:\n\\[ \\text{Area} = \\frac{1}{2} \\times \\text{base} \\times \\text{height} \\]\n\nThe base of triangle $ABC$ is $BC = 8\\sqrt{6}$.\n\nThe height of triangle $ABC$ is the distance from $B$ to side $AC$.\n\nThe area of triangle $ABC$ is:\n\\[ \\text{Area} = \\frac{1}{2} \\times \\text{base} \\times \\text{height} = \\frac{1}{2} \\times 8\\sqrt{6} \\times \\text{distance} \\]\n\nThe distance from the orthocenter to vertex $B$ is:\n\\[ \\text{distance} = \\frac{1}{2} \\times \\text{base} \\times \\text{height} = \\frac{1}{2} \\times 8\\sqrt{6} \\times \\text{distance} \\]\n\nThe distance from the orthocenter to vertex $B$ is:\n\\[ \\boxed{\\frac{1}{2} \\times 8\\sqrt{6}} \\]\nFinal answer:\n#### \\frac{1}{2} \\times 8\\sqrt{6}"}, {"kind": "sampled", "answer": "The orthocenter $H$ of a triangle is the point where the three altitudes intersect. We know that the sum of the angles in a triangle is $180^\\circ$. In this case, we can find the measure of angle $C$ first.\n\nThe sum of the angles in triangle $ABC$ is $180^\\circ$, so:\n\\[ \\alpha + \\gamma + 60^\\circ = 180^\\circ \\Rightarrow \\alpha + 75^\\circ + \\gamma = 180^\\circ \\Rightarrow \\alpha + \\gamma = 180^\\circ - 75^\\circ - 60^\\circ = 75^\\circ \\]\n\nNow, we can find the measure of angle $B$. Since the sum of angles in a triangle is $180^\\circ$, we can find the measure of angle $B$ by subtracting the sum of angles $\\alpha$ and $\\gamma$ from $180^\\circ$:\n\\[ B = 180^\\circ - (\\alpha + \\gamma) - \\alpha - \\gamma = 180^\\circ - 75^\\circ - 60^\\circ = 55^\\circ \\]\n\nThe orthocenter $H$ is the intersection of the altitudes. We can draw an altitude from vertex $B$ to side $AC$.\n\nThe distance from the orthocenter $H$ to vertex $B$ is the length of the altitude. We can find this distance by using the properties of right triangles.\n\nLet's denote the distance from $H$ to $BC$ as $x$. Then, we can use the Pythagorean theorem to find the length of the altitude:\n\\[ x^2 + 8\\sqrt{6}^2 = 25^2 \\Rightarrow x^2 = 64 \\Rightarrow x = 8 \\]\n\nSo, the distance from the orthocenter $H$ to vertex $B$ is $8$.\n\nThe answer is:\n\\[ \\boxed{8} \\]\n\nFinal answer:\n#### 8"}]} +{"index": 23, "row": {"problem": "What is the largest possible [distance](https://artofproblemsolving.com/wiki/index.php/Distance) between two [points](https://artofproblemsolving.com/wiki/index.php/Point), one on the [sphere](https://artofproblemsolving.com/wiki/index.php/Sphere) of [radius](https://artofproblemsolving.com/wiki/index.php/Radius) 19 with [center](https://artofproblemsolving.com/wiki/index.php/Center) $(-2,-10,5)$ and the other on the sphere of radius 87 with center $(12,8,-16)$?", "solution": "The distance between the two centers of the spheres can be determined via the [distance formula](https://artofproblemsolving.com/wiki/index.php/Distance_formula) in three dimensions: $\\sqrt{(12 - (-2))^2 + (8 - (-10))^2 + (-16 - 5)^2} = \\sqrt{14^2 + 18^2 + 21^2} = 31$. The largest possible distance would be the sum of the two radii and the distance between the two centers, making it $19 + 87 + 31 = \\boxed{137}$.", "answer": "137", "source": "amc_aime", "problem_type": "Geometry", "uuid": "a13c5053-1550-5657-9683-b42fad51a370", "difficulty_band": "medium", "correctness_count": 4}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\nWhat is the largest possible [distance](https://artofproblemsolving.com/wiki/index.php/Distance) between two [points](https://artofproblemsolving.com/wiki/index.php/Point), one on the [sphere](https://artofproblemsolving.com/wiki/index.php/Sphere) of [radius](https://artofproblemsolving.com/wiki/index.php/Radius) 19 with [center](https://artofproblemsolving.com/wiki/index.php/Center) $(-2,-10,5)$ and the other on the sphere of radius 87 with center $(12,8,-16)$?", "candidates": [{"kind": "greedy", "answer": "To find the largest possible distance between two points on the sphere, we can use the distance formula:\n\\[ d = \\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \\]\n\nHere, $(x_1, y_1, z_1) = (-2, -10, 5)$ and $(x_2, y_2, z_2) = (12, 8, -16)$.\n\nSubstitute these values into the distance formula:\n\\[ d = \\sqrt{(12 - (-2))^2 + (8 - (-10))^2 + (-16 - 5)^2} \\]\n\n\\[ d = \\sqrt{(14)^2 + (18)^2 + (-21)^2} \\]\n\n\\[ d = \\sqrt{196 + 324 + 441} \\]\n\n\\[ d = \\sqrt{961} \\]\n\n\\[ d = 31 \\]\n\nThe largest possible distance between two points on the sphere is:\n\\[ \\boxed{31} \\]"}, {"kind": "sampled", "answer": "The distance between the two points is the radius of the sphere, and the two points are on the same side of the center of the sphere.\n\nThe distance between the two points is 87 - 19 = 68 units.\n\nThe answer is:\n\\boxed{68}"}]} +{"index": 24, "row": {"problem": "Sally is thinking of a positive four-digit integer. When she divides it by any one-digit integer greater than $1$, the remainder is $1$. How many possible values are there for Sally's four-digit number?", "solution": "1. **Identify the condition for the number:**\n Sally's number, \\( N \\), when divided by any one-digit integer greater than 1, leaves a remainder of 1. This means:\n \\[\n N \\equiv 1 \\pmod{2}, \\quad N \\equiv 1 \\pmod{3}, \\quad N \\equiv 1 \\pmod{4}, \\quad N \\equiv 1 \\pmod{5}, \\quad N \\equiv 1 \\pmod{6}, \\quad N \\equiv 1 \\pmod{7}, \\quad N \\equiv 1 \\pmod{8}, \\quad N \\equiv 1 \\pmod{9}\n \\]\n\n2. **Find the least common multiple (LCM):**\n To satisfy all these congruences, \\( N-1 \\) must be a multiple of the least common multiple (LCM) of the numbers 2, 3, 4, 5, 6, 7, 8, and 9. We calculate the LCM as follows:\n \\[\n \\text{LCM}(2, 3, 4, 5, 6, 7, 8, 9)\n \\]\n Breaking down each number into its prime factors:\n \\[\n 2 = 2, \\quad 3 = 3, \\quad 4 = 2^2, \\quad 5 = 5, \\quad 6 = 2 \\times 3, \\quad 7 = 7, \\quad 8 = 2^3, \\quad 9 = 3^2\n \\]\n The LCM is found by taking the highest power of each prime that appears:\n \\[\n \\text{LCM} = 2^3 \\times 3^2 \\times 5 \\times 7 = 8 \\times 9 \\times 5 \\times 7 = 2520\n \\]\n\n3. **Formulate the general solution:**\n Since \\( N \\equiv 1 \\pmod{2520} \\), we can write:\n \\[\n N = 2520k + 1\n \\]\n where \\( k \\) is an integer.\n\n4. **Determine the range for four-digit numbers:**\n Four-digit numbers range from 1000 to 9999. We need to find the values of \\( k \\) such that:\n \\[\n 1000 \\leq 2520k + 1 \\leq 9999\n \\]\n Solving for \\( k \\):\n \\[\n 999 \\leq 2520k \\leq 9998\n \\]\n \\[\n \\frac{999}{2520} \\leq k \\leq \\frac{9998}{2520}\n \\]\n \\[\n 0.396 \\leq k \\leq 3.968\n \\]\n Since \\( k \\) must be an integer, the possible values for \\( k \\) are 1, 2, and 3.\n\n5. **Calculate the corresponding values of \\( N \\):**\n \\[\n k = 1 \\implies N = 2520 \\times 1 + 1 = 2521\n \\]\n \\[\n k = 2 \\implies N = 2520 \\times 2 + 1 = 5041\n \\]\n \\[\n k = 3 \\implies N = 2520 \\times 3 + 1 = 7561\n \\]\n\nThus, there are 3 possible values for Sally's four-digit number.\n\nThe final answer is \\(\\boxed{3}\\)", "answer": "3", "source": "aops_forum", "problem_type": "Number Theory", "uuid": "16a20bb8-7a4b-5b56-953e-ee73a3a59065", "difficulty_band": "medium", "correctness_count": 2}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\nSally is thinking of a positive four-digit integer. When she divides it by any one-digit integer greater than $1$, the remainder is $1$. How many possible values are there for Sally's four-digit number?", "candidates": [{"kind": "greedy", "answer": "To solve this problem, we need to find all the possible values of Sally's four-digit number that satisfy the given conditions.\n\nFirst, let's consider the condition that when Sally divides her four-digit number by any one-digit integer greater than $1$, the remainder is $1$. This means that the number must be of the form $1000k + 1$, where $k$ is a positive integer.\n\nNow, let's consider the condition that the remainder when Sally divides her four-digit number by any one-digit integer greater than $1$ is $1$. This means that the number must be of the form $1000m + 1$, where $m$ is a positive integer.\n\nWe can write the following equations:\n\n$$1000k + 1 = 1000m + 1$$\n\nSimplifying this equation, we get:\n\n$$1000k - 1000m = 0$$\n\nFactoring out $1000$ from the left side of the equation, we get:\n\n$$1000(k - m) = 0$$\n\nSince $1000$ is not equal to $0$, we must have $k - m = 0$, which means $k = m$.\n\nNow, let's consider the condition that the remainder when Sally divides her four-digit number by any one-digit integer greater than $1$ is $1$. This means that the number must be of the form $1000m + 1$, where $m$ is a positive integer.\n\nWe can write the following equations:\n\n$$1000m + 1 = 1000m + 1$$\n\nSimplifying this equation, we get:\n\n$$1000m - 1000m = 0$$\n\nFactoring out $1000$ from the left side of the equation, we get:\n\n$$1000m(1 - m) = 0$$\n\nSince $1000$ is not equal to $0$, we must have $1 - m = 0$, which means $m = 1$.\n\nTherefore, there is only one possible value for Sally's four-digit number.\n\nThe answer is:\n\\"}, {"kind": "sampled", "answer": "For any one-digit integer greater than $1$, the possible remainders when dividing by itself are $2$ and $3$.\nIf Sally divides the number by a one-digit integer greater than $1$, the remainder is $1$, which means she must divide the number by $3$ or $9$.\n\nThere are two possibilities:\n1. Sally divides the number by $3$.\n2. Sally divides the number by $9$.\n\nSo, Sally's four-digit number can be $3$ or $9$ (or $39$).\n\nThe answer is:\n\\[ \\boxed{2} \\]"}]} +{"index": 25, "row": {"problem": "Example 3: 10 people go to a bookstore to buy books. It is known that each person bought three types of books, and any two people have at least one book in common. Question: What is the maximum number of people who bought the most purchased book, at a minimum? (No. 8", "solution": "Let the total number of books sold be $n$, and the set of books bought by the $i$-th person be $A_{i}(i=1,2, \\cdots, 10)$. Construct a table of set element relationships, where the $i$-th row has $m_{i}$ ones.\n\nEstimate the total number of times each element appears, we have\n$$\\sum_{i=1}^{n} m_{i}=S=\\sum_{i=1}^{10}\\left|A_{i}\\right|=\\sum_{i=1}^{10} 3=30$$\n\nNext, calculate the total number of times each element appears in the intersections, we have\n$$\\sum_{i=1}^{n} \\mathrm{C}_{m_{i}}^{2}=\\sum_{1 \\leqslant i1, b$ is a positive rational number, $a^{b}+a^{-0}$ $=2 \\sqrt{2}$, find the value of $a^{b}-a^{-b}$.", "solution": "Consider the following solution:\nLet $a^{b}=x$, then $x+\\frac{1}{x}=2 \\sqrt{2}$.\nTransform it into $x^{2}-2 \\sqrt{2} x+1=0$.\nSolving yields $x_{1}=\\sqrt{2}+1, x_{2}=\\sqrt{2}-1$.\n$$\n\\text { When } \\begin{aligned}\nx & =\\sqrt{2}+1, \\\\\n& a^{b}-a^{-b} \\\\\n& =\\sqrt{2}+1-\\frac{1}{\\sqrt{2}+1} \\\\\n& =\\sqrt{2}+1-(\\sqrt{2}-1) \\\\\n& =2\n\\end{aligned}\n$$\n\nWhen $x=\\sqrt{2}-1$,\n$$\n\\begin{aligned}\n& a^{b}-a^{-b} \\\\\n= & \\sqrt{2}-1-\\frac{1}{\\sqrt{2}-1} \\\\\n= & \\sqrt{2}-1-(\\sqrt{2}+1) \\\\\n= & -2 .\n\\end{aligned}\n$$\n\nThis solution is incorrect. The error lies in neglecting the conditions $a>1$ and $b$ being a positive integer - if these conditions were considered, it would be clear that $a^{b}>1$. Therefore, the correct answer should be 2.", "answer": "2", "source": "cn_contest", "problem_type": "Algebra", "uuid": "d43bce07-3c8f-538f-b5ff-d00c66242b91", "difficulty_band": "hard", "correctness_count": 1}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\nExample 1. If $a>1, b$ is a positive rational number, $a^{b}+a^{-0}$ $=2 \\sqrt{2}$, find the value of $a^{b}-a^{-b}$.", "candidates": [{"kind": "greedy", "answer": "Since $a>1$, we know that $a^{b}>a^{-b}$.\n\nTherefore, $a^{b}-a^{-b}=(a^{b})^{1/b}-(a^{-b})^{1/b}=(a^{b})^{1/b}-(a^{-b})^{1/b}=(a^{b}-a^{-b})(a^{b}+a^{-b})=(a^{b}-a^{-b})(a^{b}+a^{-b})=(a^{b}-a^{-b})(a^{b}+a^{-b})=(a^{b}-a^{-b})(a^{b}+a^{-b})=(a^{b}-a^{-b})(a^{b}+a^{-b})=(a^{b}-a^{-b})(a^{b}+a^{-b})=(a^{b}-a^{-b})(a^{b}+a^{-b})=(a^{b}-a^{-b})(a^{b}+a^{-b})=(a^{b}-a^{-b})(a^{b}+a^{-b})=(a^{b}-a^{-b})(a^{b}+a^{-b})=(a^{b}-a^{-b})(a^{b}+a^{-b})=(a^{b}-a^{-b})(a^{b}+a^{-b})=(a^{b}-a^{-b})(a^{b}+a^{-b})=(a^{b}-a^{-b})(a^{b}+a^{-b})=(a^{b}-a^{-b})(a^{b}+a^{-b})=(a^{b}-a^{-b})(a^{b}+a^{-b})=(a^{b}-a^{-b})(a^{b}+a^{-b})=(a^{b}-a^{-b})(a^{b}+a^{-b})=(a^{b}-a^{-b})(a^{b}+a^{-b})=(a^{b}-a^{-b})(a^{b}+a^{-b})=(a^{b}-a^{-b})(a^{b}"}, {"kind": "sampled", "answer": "Since $a>1$, $b$ is a positive rational number. Let $a=2$, $b=q$, $q>1$. Then, $a^{b}+a^{-b}=2^{q}+2^{-q}=2\\cdot2^{q-1}=2^{q+q-2}=2^{2q-2}=2^{2q-2}$.\n\nSince $a^{b}-a^{-b}=2^{b}=(a^{b})^{2}-a^{-b}=(a^{b})^{2}-(a^{-b})^{2}$,\n\n$=(a^{b}+a^{-b})(a^{b}-a^{-b})=(2^{q+q-2})(2^{2q-2})=(2^{2q-2})(2^{2q-2})=(2^{2q-2})^{2}=(2^{2q-2})^{2}=2^{4q-8}$.\n\nSince $q>1$, $4q-8>0$.\n\nTherefore, $2^{4q-8}=2^{(2q-2)+2q-2}=2^{2q-2}$.\n\nTherefore, $a^{b}-a^{-b}=\\boxed{2^{2q-2}}$.\n\nFinal answer:\n#### 2"}]} +{"index": 41, "row": {"problem": "2. The ten positive integers from $1 \\sim 10$ are written in a row in some order, denoted as $a_{1}, a_{2}, \\cdots, a_{10}, S_{1}=a_{1}, S_{2}=a_{1}+a_{2}$, $\\cdots, S_{10}=a_{1}+a_{2}+\\cdots+a_{10}$. Then, among $S_{1}, S_{2}, \\cdots, S_{10}$, the maximum number of primes that can occur is .", "solution": "2. 7 .\n\nAdding an odd number changes the sum to the opposite parity, and among even numbers, only 2 is a prime. Let $b_{i}$ be the $i$-th ($i=1,2,3,4,5$) odd number in this row. Then, when adding $b_{2}$ and $b_{4}$, the sums $S_{k}$ and $S_{n}$ are even numbers greater than 2. Therefore, $S_{k} \\backslash S_{n}$ and $S_{10}=55$ must be composite numbers, meaning that in $S_{1}, S_{2}, \\cdots, S_{10}$, the primes are no greater than 7. The example in Table 1 shows that there can be 7 primes.\nTable 1\n\\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|}\n\\hline$i$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\\\\n\\hline$a_{i}$ & 7 & 6 & 4 & 2 & 9 & 1 & 3 & 5 & 10 & 8 \\\\\n\\hline$S_{i}$ & 7 & 13 & 17 & 19 & 28 & 29 & 32 & 37 & 47 & 55 \\\\\n\\hline\n\\end{tabular}\n\nTherefore, in $S_{1}, S_{2}, \\cdots, S_{10}$, there can be at most 7 primes.", "answer": "7", "source": "cn_contest", "problem_type": "Number Theory", "uuid": "0dedef0f-9dd7-5da7-adce-933a42bd4de6", "difficulty_band": "hard", "correctness_count": 1}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\n2. The ten positive integers from $1 \\sim 10$ are written in a row in some order, denoted as $a_{1}, a_{2}, \\cdots, a_{10}, S_{1}=a_{1}, S_{2}=a_{1}+a_{2}$, $\\cdots, S_{10}=a_{1}+a_{2}+\\cdots+a_{10}$. Then, among $S_{1}, S_{2}, \\cdots, S_{10}$, the maximum number of primes that can occur is .", "candidates": [{"kind": "greedy", "answer": "2. The ten positive integers from $1 \\sim 10$ are written in a row in some order, denoted as $a_{1}, a_{2}, \\cdots, a_{10}, S_{1}=a_{1}, S_{2}=a_{1}+a_{2}$, $\\cdots, S_{10}=a_{1}+a_{2}+\\cdots+a_{10}$.\n\nThe number of primes in the sequence $S_{1}, S_{2}, \\cdots, S_{10}$ is the maximum number of primes that can occur in the sequence $S_{1}, S_{2}, \\cdots, S_{10}$.\n\nThe maximum number of primes that can occur in the sequence $S_{1}, S_{2}, \\cdots, S_{10}$ is 10.\n\nTherefore, the answer is $\\boxed{10}$.\n\nThis problem tests the understanding of the properties of sequences and the ability to reason and calculate. It is a moderately difficult problem."}, {"kind": "sampled", "answer": "2. The numbers $S_{2}$ to $S_{10}$ are filled from $1$ to $10$. We are given that $a_{1}, a_{2}, \\cdots, a_{10} \\in S_{2,1} = S_{1}$, $S_{2}, S_{3}, \\cdots, S_{10} \\in S_{3,1}, S_{4}, \\cdots, S_{10} \\in S_{4,1}, S_{5}, \\cdots, S_{10} \\in S_{5,1}, \\cdots, S_{10} \\in S_{6,1}, S_{7}, \\cdots, S_{10} \\in S_{7,1}, S_{8}, \\cdots, S_{10} \\in S_{8,1}, S_{9}, \\cdots, S_{10} \\in S_{9,1}, S_{10}$, and $S_{10} = S_{1}$. When we compare $S_{10}$, we can see that $a_{1} \\in S_{10}$ is the maximum.\n\nFinal answer:\n#### 6"}]} +{"index": 42, "row": {"problem": "4. Given arithmetic sequences $\\left\\{a_{n}\\right\\},\\left\\{b_{n}\\right\\}$, the sums of the first $n$ terms are $S_{n}, T_{n}$ respectively, and $\\frac{S_{n}}{T_{n}}=\\frac{3 n+2}{2 n+1}$. Then $\\frac{a_{7}}{b_{5}}=$ $\\qquad$", "solution": "4. $\\frac{41}{19}$.\n\nFrom $\\frac{S_{n}}{T_{n}}=\\frac{3 n+2}{2 n+1}$, we get $\\frac{S_{n}}{T_{n}}=\\frac{\\left(3 n^{2}+2 n\\right) k}{\\left(2 n^{2}+n\\right) k}$.\nLet $S_{n}=\\left(3 n^{2}+2 n\\right) k, T_{n}=\\left(2 n^{2}+n\\right) k$, then\n$$\na_{7}=S_{7}-S_{6}=41 k, \\quad b_{5}=T_{5}-T_{4}=19 k,\n$$\n\nThus $\\frac{a_{7}}{b_{5}}=\\frac{41 k}{19 k}=\\frac{41}{19}$.", "answer": "\\frac{41}{19}", "source": "olympiads", "problem_type": "Algebra", "uuid": "f148d61e-86c5-5387-83e9-083cf347e443", "difficulty_band": "medium", "correctness_count": 2}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\n4. Given arithmetic sequences $\\left\\{a_{n}\\right\\},\\left\\{b_{n}\\right\\}$, the sums of the first $n$ terms are $S_{n}, T_{n}$ respectively, and $\\frac{S_{n}}{T_{n}}=\\frac{3 n+2}{2 n+1}$. Then $\\frac{a_{7}}{b_{5}}=$ $\\qquad$", "candidates": [{"kind": "greedy", "answer": "1. **Given Information and Initial Setup:**\n - The arithmetic sequences $\\left\\{a_{n}\\right\\}$ and $\\left\\{b_{n}\\right\\}$ have sums $S_{n}$ and $T_{n}$ respectively.\n - The ratio $\\frac{S_{n}}{T_{n}} = \\frac{3n + 2}{2n + 1}$.\n\n2. **Expressing $S_{n}$ and $T_{n}$ in Terms of $a_{1}$ and $d$:**\n - For the arithmetic sequence $\\left\\{a_{n}\\right\\}$, we have $S_{n} = na_{1} + \\frac{n(n-1)}{2}d$.\n - For the arithmetic sequence $\\left\\{b_{n}\\right\\}$, we have $T_{n} = b_{1} + \\frac{n(n-1)}{2}d$.\n\n3. **Substituting $S_{n}$ and $T_{n}$ into the Given Ratio:**\n - The ratio $\\frac{S_{n}}{T_{n}} = \\frac{na_{1} + \\frac{n(n-1)}{2}d}{b_{1} + \\frac{n(n-1)}{2}d}$.\n\n4. **Simplifying the Ratio:**\n - Simplify the ratio by canceling out the common terms in the numerator and denominator.\n - This gives us $\\frac{S_{n}}{T_{n}} = \\frac{a_{1}}{b_{1}} + \\frac{n(n-1)}{2}d$.\n\n5. **Using the Given Ratio:**\n - Given $\\frac{S_{n}}{T_{n}} = \\frac{3n + 2}{2n + 1}$, we can equate the two expressions.\n - This leads to $\\frac{a_{1}}{b_{1}} + \\frac{n(n-1)}{2}d = \\frac{3n + 2}{2n + 1}$.\n\n6. **Solving for $\\frac{a_{1}}{b_{1}}$:**\n - Rearrange the equation to isolate $\\frac{a_{1}}{b_{1}}$.\n - This gives us $\\frac{a_{1}"}, {"kind": "sampled", "answer": "1. We start by analyzing the relationship between the arithmetic sequences $\\{a_{n}\\}$ and $\\{b_{n}\\}$. Let the common difference of the arithmetic sequence $\\{a_{n}\\}$ be denoted by $d$. We know that $a_{n} = a_{1} + (n-1)d$ and $b_{n} = b_{1} + (n-1)d$.\n\n\\[ a_{n} = a_{1} + (n-1)d \\]\n\\[ b_{n} = b_{1} + (n-1)d \\]\n\n2. The sum of the first $n$ terms of an arithmetic sequence is given by:\n\n\\[ S_{n} = \\frac{n}{2} [2a_{1} + (n-1)d] \\]\n\\[ T_{n} = \\frac{n}{2} [2b_{1} + (n-1)d] \\]\n\n3. We are given the ratio of the sums:\n\n\\[ \\frac{S_{n}}{T_{n}} = \\frac{3n + 2}{2n + 1} \\]\n\n4. Substitute the expressions for $S_{n}$ and $T_{n}$ into the equation:\n\n\\[ \\frac{\\frac{n}{2} [2a_{1} + (n-1)d]}{\\frac{n}{2} [2b_{1} + (n-1)d]} = \\frac{3n + 2}{2n + 1} \\]\n\n5. Simplify the equation by canceling out the common factor of $\\frac{n}{2}$:\n\n\\[ \\frac{2a_{1} + (n-1)d}{\\frac{n}{2} [2b_{1} + (n-1)d]} = \\frac{3n + 2}{2n + 1} \\]\n\n6. Cross-multiply to clear the fractions:\n\n\\[ (2a_{1} + (n-1)d) \\left(2n + 1\\right) = n[2b_{1} + (n-1)d] \\left(2n + 2\\right) \\]\n\n7. Expanding both sides of the equation:"}]} +{"index": 43, "row": {"problem": "Find all natural numbers $n> 1$ for which the following applies:\nThe sum of the number $n$ and its second largest divisor is $2013$.\n\n(R. Henner, Vienna)", "solution": "1. Let \\( n \\) be a natural number greater than 1, and let \\( p \\) be the smallest prime divisor of \\( n \\). The second largest divisor of \\( n \\) is then \\( \\frac{n}{p} \\).\n2. According to the problem, the sum of \\( n \\) and its second largest divisor is 2013. Therefore, we have:\n \\[\n n + \\frac{n}{p} = 2013\n \\]\n3. Factor out \\( n \\) from the left-hand side:\n \\[\n n \\left(1 + \\frac{1}{p}\\right) = 2013\n \\]\n4. Simplify the expression inside the parentheses:\n \\[\n n \\left(\\frac{p+1}{p}\\right) = 2013\n \\]\n5. Solve for \\( n \\):\n \\[\n n = \\frac{2013p}{p+1}\n \\]\n6. Since \\( n \\) must be a natural number, \\( \\frac{2013p}{p+1} \\) must also be a natural number. This implies that \\( p+1 \\) must be a divisor of 2013.\n7. Factorize 2013:\n \\[\n 2013 = 3 \\times 11 \\times 61\n \\]\n The divisors of 2013 are: 1, 3, 11, 33, 61, 183, 671, and 2013.\n8. We need \\( p+1 \\) to be one of these divisors. Therefore, we check the possible values for \\( p \\):\n \\[\n p+1 = 3 \\implies p = 2\n \\]\n \\[\n p+1 = 11 \\implies p = 10 \\quad (\\text{not a prime})\n \\]\n \\[\n p+1 = 61 \\implies p = 60 \\quad (\\text{not a prime})\n \\]\n The only valid prime \\( p \\) is 2.\n9. Substitute \\( p = 2 \\) back into the equation for \\( n \\):\n \\[\n n = \\frac{2013 \\times 2}{2+1} = \\frac{4026}{3} = 1342\n \\]\n10. Verify the solution:\n \\[\n 1342 + \\frac{1342}{2} = 1342 + 671 = 2013\n \\]\n The condition is satisfied.\n\nThe final answer is \\( \\boxed{ n = 1342 } \\).", "answer": "n = 1342", "source": "aops_forum", "problem_type": "Number Theory", "uuid": "baf55b5c-1654-5be9-acf7-1f553e23ffc7", "difficulty_band": "medium", "correctness_count": 2}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\nFind all natural numbers $n> 1$ for which the following applies:\nThe sum of the number $n$ and its second largest divisor is $2013$.\n\n(R. Henner, Vienna)", "candidates": [{"kind": "greedy", "answer": "To find all natural numbers $n>1$ for which the sum of the number $n$ and its second largest divisor is $2013$, we need to consider the properties of numbers and their divisors.\n\nFirst, let's recall that a number $n$ has a divisor $d$ if $n$ is divisible by $d$. Therefore, the second largest divisor of $n$ is $d$.\n\nNow, let's consider the sum of $n$ and its second largest divisor. We can express this sum as:\n\\[ n + d = 2013 \\]\n\nSince $n$ is a natural number greater than $1$, we can start by trying small values of $n$ and their second largest divisor.\n\nFor $n = 2$, the second largest divisor is $d = 1$. However, $2 + 1 = 3$, which does not satisfy the equation.\n\nFor $n = 3$, the second largest divisor is $d = 1$. However, $3 + 1 = 4$, which does not satisfy the equation.\n\nFor $n = 4$, the second largest divisor is $d = 1$. However, $4 + 1 = 5$, which does not satisfy the equation.\n\nFor $n = 5$, the second largest divisor is $d = 1$. However, $5 + 1 = 6$, which does not satisfy the equation.\n\nFor $n = 6$, the second largest divisor is $d = 1$. However, $6 + 1 = 7$, which does not satisfy the equation.\n\nFor $n = 7$, the second largest divisor is $d = 1$. However, $7 + 1 = 8$, which does not satisfy the equation.\n\nFor $n = 8$, the second largest divisor is $d = 1$. However, $8 + 1 = 9$, which does not satisfy the equation.\n\nFor $n = 9$, the second largest divisor is $d = 1$. However, $9 + 1 = 10$, which does not satisfy the equation.\n\nFor $n = 10$, the second largest divisor is $d = 1$. However, $10 + 1 = 11"}, {"kind": "sampled", "answer": "First, we want to find a general form for every $n > 1$.\n\nLet $n$ be a natural number. Then the second largest divisor $d$ of $n$ is $1 + 1 + 2 + \\dots + (n-1)$.\n\nSo, $S(n) = n(n-1) + (n-1) + \\dots + 2 = 2013$.\n\nThis means that $S(n)$ is an odd number.\n\nTherefore, $n$ must be odd.\n\nSince $n>1$, the next possible values for $n$ are $n=3, 5, 7, 9, 11, \\dots$.\n\nFinally, there are no other values of $n$ to consider.\n\nAnswer:\n#### 2"}]} +{"index": 44, "row": {"problem": "99(1209). How many consecutive natural numbers, starting from 1, need to be added to get a three-digit number written with identical digits?", "solution": "Solution. Let $1+2+3+\\ldots+n=\\overline{b b b}$. Then $S_{n}=$ $=\\frac{(1+n) \\cdot n}{2}$. We have $\\frac{n(n+1)}{2}=100 b+10 b+b$, from which\n\n$$\nn(n+1)=2 \\cdot 111 b, n(n+1)=2 \\cdot 3 \\cdot 37 b=6 b \\cdot 37\n$$\n\nwhere $n<45(111 b<1000,222 b<2000, n<45)$.\n\nSince the left side of the equation is the product of two consecutive natural numbers, the right side of the equation must also be the product of two consecutive natural numbers. One of them is prime (37), and the other is a multiple of 6, so it must be 36. We have $n(n+1)=36 \\cdot 37, n=36$.\n\nIt is advisable to check:\n\n$$\n1+2+\\ldots+36=\\frac{(1+36) \\cdot 36}{2}=666\n$$\n\nAnswer: 36 numbers need to be added.", "answer": "36", "source": "olympiads", "problem_type": "Number Theory", "uuid": "10dd6b04-7a5c-5dfe-9172-6272962efcf2", "difficulty_band": "medium", "correctness_count": 2}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\n99(1209). How many consecutive natural numbers, starting from 1, need to be added to get a three-digit number written with identical digits?", "candidates": [{"kind": "greedy", "answer": "99(1209) = 11(1209) = 1339\n1339 + 1 = 1340\n\nFinal answer:\n#### 1340"}, {"kind": "sampled", "answer": "To solve this, we need to find how many consecutive natural numbers from 1 to 99 have identical digits and, when we add these numbers, the result is a three-digit number with identical digits.\n\nHere's the step-by-step solution:\n\n1. Identify the smallest and largest number in the range, which are both three-digit numbers with identical digits.\n\n2. The range starts from 1 and ends at 99, with a length of 99 - 1 + 1 = 99 + 1 = 100.\n\n3. The range includes 1 to 99.\n\n4. For the three-digit numbers, the range includes:\n - 1 to 99: 99 * 100 = 9900 (since 100 + 99 = 199)\n - 100 to 999: 99 * 1000 = 99000\n - 1000 to 9999: 99 * 10000 = 990000.\n\nIn summary, the range includes 9900 consecutive natural numbers, starting from 1 and ending at 99000, which is a total of 99000 three-digit numbers. When we add these numbers, we get a three-digit number with identical digits, which is a number like 1001.\n\nFinal answer:\n#### 1001"}]} +{"index": 45, "row": {"problem": "B1. Let $A B C D$ be a trapezoid with bases $A B$ and $C D$. Let $|A B|=6.5 \\, \\text{cm}, |B C|=4.3 \\, \\text{cm}$, $|A D|=3.8 \\, \\text{cm}$, and $|A C|=5.3 \\, \\text{cm}$. Calculate the size of the angle $\\angle C B A$, the height of the trapezoid $A B C D$, and its area.", "solution": "B1. Let $a=|A B|, b=|B C|, c=|C D|, d=|A D|, e=|A C|$. Let $X$ and $Y$ be the feet of the altitudes from $C$ and $D$ respectively, and let $x=|X B|$ and $y=|A Y|$. Using the cosine rule, we calculate the angle $\\cos \\beta=\\frac{a^{2}+b^{2}-e^{2}}{2 a b} \\Rightarrow \\beta \\doteq 54.26^{\\circ} \\doteq 54^{\\circ} 16'$. In the right triangle with sides $|X C|,|X B|$, and $|B C|$ and angle $\\beta$, we calculate the height of the trapezoid $v=b \\cdot \\sin \\beta \\doteq 3.5 \\mathrm{~cm}$ and the side $x=\\sqrt{b^{2}-v^{2}} \\doteq 2.5 \\mathrm{~cm}$. In the other right triangle with sides $|A Y|,|Y D|$, and $|A D|$, using the Pythagorean theorem, we calculate the side $y=\\sqrt{d^{2}-v^{2}} \\doteq 1.5 \\mathrm{~cm}$ and thus obtain the length of the side $c=a-x-y \\doteq 2.5 \\mathrm{~cm}$. Using the formula $S=\\frac{a+c}{2} \\cdot v$, we calculate the area of the trapezoid $S \\doteq 15.75 \\mathrm{~cm}^{2}$.\n\n![](https://cdn.mathpix.com/cropped/2024_06_07_9fbd0ae098c18b73b6bfg-24.jpg?height=55&width=1739&top_left_y=1503&top_left_x=176)\n\n![](https://cdn.mathpix.com/cropped/2024_06_07_9fbd0ae098c18b73b6bfg-24.jpg?height=48&width=1739&top_left_y=1552&top_left_x=176)\n\n![](https://cdn.mathpix.com/cropped/2024_06_07_9fbd0ae098c18b73b6bfg-24.jpg?height=51&width=1739&top_left_y=1599&top_left_x=176)\n\n![](https://cdn.mathpix.com/cropped/2024_06_07_9fbd0ae098c18b73b6bfg-24.jpg?height=57&width=1742&top_left_y=1645&top_left_x=174)\n\n![](https://cdn.mathpix.com/cropped/2024_06_07_9fbd0ae098c18b73b6bfg-24.jpg?height=66&width=1736&top_left_y=1692&top_left_x=177)\n\n![](https://cdn.mathpix.com/cropped/2024_06_07_9fbd0ae098c18b73b6bfg-24.jpg?height=52&width=1736&top_left_y=1753&top_left_x=177)\nThe formula for the area $S=\\frac{a+c}{2} \\cdot v \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots . \\ldots . \\ldots . \\ldots$ point The calculated area of the trapezoid $S \\doteq 15.75 \\mathrm{~cm}^{2} \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots$", "answer": "15.75\\,^2", "source": "olympiads", "problem_type": "Geometry", "uuid": "f975a686-9d23-5d7b-a68d-9e96c861088f", "difficulty_band": "hard", "correctness_count": 1}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\nB1. Let $A B C D$ be a trapezoid with bases $A B$ and $C D$. Let $|A B|=6.5 \\, \\text{cm}, |B C|=4.3 \\, \\text{cm}$, $|A D|=3.8 \\, \\text{cm}$, and $|A C|=5.3 \\, \\text{cm}$. Calculate the size of the angle $\\angle C B A$, the height of the trapezoid $A B C D$, and its area.", "candidates": [{"kind": "greedy", "answer": "The area of a trapezoid is given by the formula $S_{\\text{trapezoid}} = \\frac{1}{2} (a + b) h$, where $a$ and $b$ are the lengths of the two parallel sides, and $h$ is the height.\n\nWe can find the height of the trapezoid by using the given values for the lengths of the bases and the height.\n\nFirst, we need to find the length of the base $a$ and the length of the base $b$. We can do this by adding the lengths of the two parallel sides.\n\n$$a = |A B| + |B C|$$\n$$b = |A D| + |C D|$$\n\nNow, plug in the given values for the lengths of the bases and the height to calculate the lengths of the bases and the height.\n\n$$a = 6.5 \\, \\text{cm} + 4.3 \\, \\text{cm}$$\n$$b = 3.8 \\, \\text{cm} + 5.3 \\, \\text{cm}$$\n\nCalculate the lengths of the bases and the height.\n\n$$a = 6.5 + 4.3 = 10.8 \\, \\text{cm}$$\n$$b = 3.8 + 5.3 = 9.1 \\, \\text{cm}$$\n\nNow that we have the lengths of the bases and the height, we can find the area of the trapezoid. This can be done by multiplying the lengths of the bases and the height by the ratio of the lengths of the bases to the height.\n\n$$S_{\\text{trapezoid}} = a h + b h$$\n\nSubstitute the values for the lengths of the bases and the height to calculate the area of the trapezoid.\n\n$$S_{\\text{trapezoid}} = 10.8 \\, \\text{cm} \\times 3.1 \\, \\text{cm} + 9.1 \\, \\text{cm} \\times 3.1 \\, \\text{cm}$$\n\nCalculate the area of the trapezoid.\n\n$$S_{\\text{trapezoid}} = 10.8 \\times 3.1 + 9.1 \\times"}, {"kind": "sampled", "answer": "From the question we know that $|A B|=6.5\\,cm$, $|A D|=3.8\\,cm$, $|A C|=5.3\\,cm$ and $|A B|=6.5\\,cm$.\n\nWe can calculate the length of the base $|A D|$ and the height $|A B|$ of the trapezoid $A B C D$.\n\n\\[ \\Rightarrow |A D| = \\frac{A B}{3} = \\frac{6.5\\,cm}{3} = \\frac{13}{6}\\,cm \\]\n\nSimilarly, we can calculate the length of the base $|A C|$ and the height $|A B|$ of the trapezoid $A B C D$.\n\n\\[ \\Rightarrow |A C| = \\frac{A B}{2} = \\frac{6.5\\,cm}{2} = \\frac{13}{4}\\,cm \\]\n\nNow, we can calculate the length of the height $|A B|$ of the trapezoid $A B C D$.\n\n\\[ \\Rightarrow |A B| = \\frac{13}{6}\\,cm + \\frac{13}{4}\\,cm = \\frac{13}{6}\\,cm + \\frac{13}{6}\\,cm = \\frac{26}{6}\\,cm = \\frac{13}{3}\\,cm \\]\n\nNow, we can calculate the area of the trapezoid $A B C D$.\n\n\\[ \\text{Area} = \\frac{1}{2} \\times (\\text{Base}_1 + \\text{Base}_2) \\times \\text{Height} = \\frac{1}{2} \\times (6.5 + 13) \\times \\frac{13}{3} = \\frac{1}{2} \\times (20) \\times \\frac{13}{3} = 26\\,cm^2 \\]\n\nThe area of the trapezoid is $26\\,cm^2$.\n\nFinal answer:\n#### 26"}]} +{"index": 46, "row": {"problem": "11. The blackboard is written with 1989 consecutive natural numbers $1, 2, 3, \\ldots, 1989$. First, perform the following transformation: erase any two numbers on the blackboard, and add the remainder obtained by dividing the sum of the erased two numbers by 19. After several transformations, there are two numbers left on the blackboard, one of which is 89, and the other is a single-digit number. This single-digit number is ( ).", "solution": "【Answer】2\n【Analysis】Key point: Number Theory, Remainder Properties. According to the property that the remainder of a sum equals the remainder of the sum of the remainders, $(1+2+3+\\ldots . .+1989) \\div 19=104160 \\ldots \\ldots .15$; $89 \\div 19=4 \\ldots \\ldots 13$; thus, the remainder when the remaining number is divided by 19 must be 2, and since it is a single digit, it can only be 2", "answer": "2", "source": "olympiads", "problem_type": "Number Theory", "uuid": "178d40c7-2ccf-5475-af03-6287e36f345c", "difficulty_band": "medium", "correctness_count": 2}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\n11. The blackboard is written with 1989 consecutive natural numbers $1, 2, 3, \\ldots, 1989$. First, perform the following transformation: erase any two numbers on the blackboard, and add the remainder obtained by dividing the sum of the erased two numbers by 19. After several transformations, there are two numbers left on the blackboard, one of which is 89, and the other is a single-digit number. This single-digit number is ( ).", "candidates": [{"kind": "greedy", "answer": "12. The problem can be solved by using the following method:\n\n$$1989 = 1 + 2 + 3 + \\ldots + 1989$$\n\n$$1989 = 1990$$\n\n$$1990 = 2 + 3 + 4 + \\ldots + 1989$$\n\n$$1990 = 2000$$\n\n$$1990 = 2001$$\n\n$$1990 = 2002$$\n\n$$1990 = 2003$$\n\n$$1990 = 2004$$\n\n$$1990 = 2005$$\n\n$$1990 = 2006$$\n\n$$1990 = 2007$$\n\n$$1990 = 2008$$\n\n$$1990 = 2009$$\n\n$$1990 = 2010$$\n\n$$1990 = 2011$$\n\n$$1990 = 2012$$\n\n$$1990 = 2013$$\n\n$$1990 = 2014$$\n\n$$1990 = 2015$$\n\n$$1990 = 2016$$\n\n$$1990 = 2017$$\n\n$$1990 = 2018$$\n\n$$1990 = 2019$$\n\n$$1990 = 2020$$\n\n$$1990 = 2021$$\n\n$$1990 = 2022$$\n\n$$1990 = 2023$$\n\n$$1990 = 2024$$\n\n$$1990 = 2025$$\n\n$$1990 = 2026$$\n\n$$1990 = 2027$$\n\n$$1990 = 2028$$\n\n$$1990 = 2029$$\n\n$$1990 = 2"}, {"kind": "sampled", "answer": "1. **Initial Setup**: The blackboard consists of 1989 consecutive natural numbers, starting from 1.\n\n2. **First Transformation**: We start by erasing any two numbers on the board. This operation will perform a series of operations until only one number remains on the board.\n\n3. **Second Transformation**: In the next step, we remove the sum of the two numbers we just erased. We do this by dividing the sum of the two remaining numbers by 19. This transformation continues until only one number remains.\n\n4. **Final Grouping**: After the second transformation, we divide the sum of the two remaining numbers by 19 until only one number remains. This process continues until only one number remains in the final group.\n\nBy following this process, we can achieve the desired result.\n\n#### 1990. **Final"}]} +{"index": 47, "row": {"problem": "Let $w_1, w_2, \\dots, w_n$ be [complex numbers](https://artofproblemsolving.com/wiki/index.php/Complex_number). A line $L$ in the [complex plane](https://artofproblemsolving.com/wiki/index.php/Complex_plane) is called a mean [line](https://artofproblemsolving.com/wiki/index.php/Line) for the [points](https://artofproblemsolving.com/wiki/index.php/Point) $w_1, w_2, \\dots, w_n$ if $L$ contains points (complex numbers) $z_1, z_2, \\dots, z_n$ such that\n\\[\\sum_{k = 1}^n (z_k - w_k) = 0.\\]\nFor the numbers $w_1 = 32 + 170i$, $w_2 = - 7 + 64i$, $w_3 = - 9 + 200i$, $w_4 = 1 + 27i$, and $w_5 = - 14 + 43i$, there is a unique mean line with $y$-intercept 3. Find the [slope](https://artofproblemsolving.com/wiki/index.php/Slope) of this mean line.", "solution": "Solution 1\n$\\sum_{k=1}^5 z_k - \\sum_{k=1}^5 w_k = 0$\n$\\sum_{k=1}^5 z_k = 3 + 504i$\nEach $z_k = x_k + y_ki$ lies on the complex line $y = mx + 3$, so we can rewrite this as \n$\\sum_{k=1}^5 z_k = \\sum_{k=1}^5 x_k + \\sum_{k=1}^5 y_ki$\n$3 + 504i = \\sum_{k=1}^5 x_k + i \\sum_{k=1}^5 (mx_k + 3)$\nMatching the real parts and the imaginary parts, we get that $\\sum_{k=1}^5 x_k = 3$ and $\\sum_{k=1}^5 (mx_k + 3) = 504$. Simplifying the second summation, we find that $m\\sum_{k=1}^5 x_k = 504 - 3 \\cdot 5 = 489$, and substituting, the answer is $m \\cdot 3 = 489 \\Longrightarrow m = 163$.\n\nSolution 2\nWe know that \n$\\sum_{k=1}^5 w_k = 3 + 504i$\nAnd because the sum of the 5 $z$'s must cancel this out,\n$\\sum_{k=1}^5 z_k = 3 + 504i$\nWe write the numbers in the form $a + bi$ and we know that \n$\\sum_{k=1}^5 a_k = 3$ and $\\sum_{k=1}^5 b_k = 504$\nThe line is of equation $y=mx+3$. Substituting in the polar coordinates, we have $b_k = ma_k + 3$.\nSumming all 5 of the equations given for each $k$, we get \n$504 = 3m + 15$\nSolving for $m$, the slope, we get $\\boxed{163}$\n\nSolution 3\nThe mean line for $w_1, . . ., w_5$ must pass through the mean (the center of mass) of these points, which, if we graph them on the complex plane, is $(\\frac{3}{5}, \\frac{504i}{5})$. Since we now have two points, namely that one and $(0, 3i)$, we can simply find the slope between them, which is $\\boxed{163}$ by the good ol' slope formula.", "answer": "163", "source": "amc_aime", "problem_type": "Algebra", "uuid": "5b7f6119-c972-53e8-8eac-e22f8ad8f87e", "difficulty_band": "medium", "correctness_count": 2}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\nLet $w_1, w_2, \\dots, w_n$ be [complex numbers](https://artofproblemsolving.com/wiki/index.php/Complex_number). A line $L$ in the [complex plane](https://artofproblemsolving.com/wiki/index.php/Complex_plane) is called a mean [line](https://artofproblemsolving.com/wiki/index.php/Line) for the [points](https://artofproblemsolving.com/wiki/index.php/Point) $w_1, w_2, \\dots, w_n$ if $L$ contains points (complex numbers) $z_1, z_2, \\dots, z_n$ such that\n\\[\\sum_{k = 1}^n (z_k - w_k) = 0.\\]\nFor the numbers $w_1 = 32 + 170i$, $w_2 = - 7 + 64i$, $w_3 = - 9 + 200i$, $w_4 = 1 + 27i$, and $w_5 = - 14 + 43i$, there is a unique mean line with $y$-intercept 3. Find the [slope](https://artofproblemsolving.com/wiki/index.php/Slope) of this mean line.", "candidates": [{"kind": "greedy", "answer": "The mean line has a slope of $\\frac{-14 + 43i}{-14 + 27i} = \\frac{17 + 3i}{3 + 9i}$.\n\nThe answer is:\n\\[ \\boxed{\\frac{17 + 3i}{3 + 9i}} \\]"}, {"kind": "sampled", "answer": "Assumptions\n\n1. The new real numbers are given as $z_1 = 170 + 43i$ and $z_2 = -7 + 64i$.\n2. The new complex numbers are given as $w_1 = 32 + 170i$ and $w_2 = -7 + 64i$.\n3. The new complex numbers are given as $w_3 = 9 + 200i$ and $w_4 = 1 + 27i$.\n4. The new complex numbers are given as $w_5 = -14 + 43i$ and $w_5' = 1 + 27i$.\n5. We are asked to find the slope of the mean line $y = mx + b$, where $m$ is the slope.\n\nFirst, we need to find the slope $m$ of the new mean line. This can be done by taking the derivative of the given mean line with respect to $x$ and setting it equal to 0.\n\nLet $m = y'$. Then:\n\\[ m = \\frac{d}{dx} (z_{1} - w_{1}) + \\frac{d}{dx} (z_{2} - w_{2}) + \\frac{d}{dx} (z_{3} - w_{3}) + \\frac{d}{dx} (z_{4} - w_{4}) + \\frac{d}{dx} (w_{5} - w_{5'}) \\]\n\nUsing the chain rule, we get:\n\\[ m = -(z_{1} - w_{1}) + (z_{2} - w_{2}) + (z_{3} - w_{3}) + (z_{4} - w_{4}) + (w_{5} - w_{5}) \\]\n\nWe can rewrite this as:\n\\[ m = -(z_{1} - w_{1}) + (z_{2} - w_{2}) + (z_{3} - w_{3}) + (z_{4} - w_{4}) + (w_{5} - w_{5'}) = 0 \\]\n\nNow, we can substitute the given values for $w_{1}$, $w_{2}$, $w_{3"}]} +{"index": 48, "row": {"problem": "In a circle with a radius of 1.4, determine the distance from the center to the chord if it intercepts an arc of $120^{\\circ}$.\n\n#", "solution": "The leg opposite the $30^{\\circ}$ angle is half the hypotenuse.\n\n## Solution\n\nLet $M$ be the foot of the perpendicular dropped from the center $O$ to the chord $A B$. Then $O M$ is the leg of the right triangle $O M A$, lying opposite the angle of $30^{\\circ}$. Therefore, $O M=1 / 2 O A=0.7$.\n\n## Answer\n\n0.7.", "answer": "0.7", "source": "olympiads", "problem_type": "Geometry", "uuid": "c0fc7493-a0af-5a11-8922-ea00d225b05c", "difficulty_band": "medium", "correctness_count": 2}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\nIn a circle with a radius of 1.4, determine the distance from the center to the chord if it intercepts an arc of $120^{\\circ}$.\n\n#", "candidates": [{"kind": "greedy", "answer": "The distance from the center of the circle to the chord is equal to the radius of the circle.\n\nFinal answer:\n#### 1.4"}, {"kind": "sampled", "answer": "Since the circle is inscribed in a line, the chord is the diameter of the circle. Therefore, the distance from the center of the circle to the chord is equal to the radius of the circle.\n\n#### 1. Define the radius of the circle.\nThe radius of the circle is given as $r = 1.4$.\n\n#### 2. Consider the triangle formed by the center of the circle, the midpoint of the chord, and the point where the chord intersects the diameter of the circle.\n\nLet the midpoint of the chord be denoted as $M$. By the problem's condition, the radius of the circle, the distance from the center of the circle to the midpoint $M$, and the point where the chord intersects the diameter of the circle form the following triangle:\n\n\\[\n\\triangle ABM = 120^{\\circ}\n\\]\n\n#### 3. Using trigonometric relationships.\nSince the radius of the circle is $r = 1.4$, the distance $AM$ can be expressed as:\n\\[\nAM = r \\sin 60^{\\circ} = 1.4 \\cdot \\frac{\\sqrt{3}}{2} = \\frac{7\\sqrt{3}}{2}\n\\]\n\n#### 4. Apply the cosine rule in $\\triangle ABM$ to find $BM$.\nWe can use the cosine rule in $\\triangle ABM$ to find $BM$:\n\\[\nBM^2 = AM^2 + AM^2 - 2 \\cdot AM \\cdot AM \\cdot \\cos 60^{\\circ}\n\\]\n\n\\[\nBM^2 = 1.4^2 + \\left(\\frac{7\\sqrt{3}}{2}\\right)^2 - 2 \\cdot 1.4 \\cdot \\frac{7\\sqrt{3}}{2} \\cdot \\cos 60^{\\circ}\n\\]\n\nSubstituting the known values for $\\cos 60^{\\circ} = \\frac{1}{2}$ and $\\sin 60^{\\circ} = \\frac{\\sqrt{3}}{2}$, we get:\n\\[\nBM^2 = 1.4^2 + \\frac{7^2}{4} - 2 \\cdot 1.4 \\cdot \\frac{\\"}]} +{"index": 49, "row": {"problem": "Basil needs to solve an exercise on summing two fractions $\\dfrac{a}{b}$ and $\\dfrac{c}{d}$, where $a$, $b$, $c$, $d$ are some non-zero real numbers. But instead of summing he performed multiplication (correctly). It appears that Basil's answer coincides with the correct answer to given exercise. Find the value of $\\dfrac{b}{a} + \\dfrac{d}{c}$.", "solution": "1. **Given Problem**: Basil needs to sum two fractions $\\frac{a}{b}$ and $\\frac{c}{d}$, but he mistakenly multiplies them. The problem states that the result of his multiplication coincides with the correct sum of the fractions.\n\n2. **Correct Sum of Fractions**: The correct sum of the fractions $\\frac{a}{b}$ and $\\frac{c}{d}$ is:\n \\[\n \\frac{a}{b} + \\frac{c}{d} = \\frac{ad + bc}{bd}\n \\]\n\n3. **Incorrect Multiplication**: Basil's multiplication of the fractions is:\n \\[\n \\frac{a}{b} \\cdot \\frac{c}{d} = \\frac{ac}{bd}\n \\]\n\n4. **Equality Condition**: According to the problem, the result of the multiplication coincides with the correct sum:\n \\[\n \\frac{ad + bc}{bd} = \\frac{ac}{bd}\n \\]\n\n5. **Simplifying the Equality**: Since the denominators are the same, we can equate the numerators:\n \\[\n ad + bc = ac\n \\]\n\n6. **Rearranging the Equation**: To find the relationship between the variables, we rearrange the equation:\n \\[\n ad + bc = ac \\implies ad + bc - ac = 0 \\implies ad - ac + bc = 0 \\implies a(d - c) + bc = 0\n \\]\n\n7. **Solving for $\\frac{b}{a} + \\frac{d}{c}$**: We need to find the value of $\\frac{b}{a} + \\frac{d}{c}$. From the equation $a(d - c) + bc = 0$, we can isolate $bc$:\n \\[\n a(d - c) = -bc \\implies \\frac{a(d - c)}{ac} = -\\frac{bc}{ac} \\implies \\frac{d - c}{c} = -\\frac{b}{a}\n \\]\n\n8. **Expressing $\\frac{d}{c}$**: Adding $\\frac{d}{c}$ to both sides:\n \\[\n \\frac{d}{c} - 1 = -\\frac{b}{a} \\implies \\frac{d}{c} = 1 - \\frac{b}{a}\n \\]\n\n9. **Summing the Fractions**: Now, we sum $\\frac{b}{a}$ and $\\frac{d}{c}$:\n \\[\n \\frac{b}{a} + \\frac{d}{c} = \\frac{b}{a} + \\left(1 - \\frac{b}{a}\\right) = \\frac{b}{a} + 1 - \\frac{b}{a} = 1\n \\]\n\nConclusion:\n\\[\n\\boxed{1}\n\\]", "answer": "1", "source": "aops_forum", "problem_type": "Algebra", "uuid": "23d8a8fb-84f9-5322-9b89-f69bc0631004", "difficulty_band": "medium", "correctness_count": 2}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\nBasil needs to solve an exercise on summing two fractions $\\dfrac{a}{b}$ and $\\dfrac{c}{d}$, where $a$, $b$, $c$, $d$ are some non-zero real numbers. But instead of summing he performed multiplication (correctly). It appears that Basil's answer coincides with the correct answer to given exercise. Find the value of $\\dfrac{b}{a} + \\dfrac{d}{c}$.", "candidates": [{"kind": "greedy", "answer": "The correct answer is 0.\n\nFinal answer:\n#### 0"}, {"kind": "sampled", "answer": "To find the value of $\\dfrac{b}{a} + \\dfrac{d}{c}$, we can use the fact that when $a = c = 1$, the expression evaluates to 1.\n\n\\[ \\frac{b}{a} + \\frac{d}{c} = \\frac{b}{1} + \\frac{d}{1} = b + d = 2 + 1 = \\boxed{3} \\]\n\nFinal answer:\n#### 3"}]} +{"index": 50, "row": {"problem": "The straight line $y=ax+16$ intersects the graph of $y=x^3$ at $2$ distinct points. What is the value of $a$?", "solution": "1. **Set up the equations:**\n The line \\( y = ax + 16 \\) intersects the curve \\( y = x^3 \\). To find the points of intersection, set the equations equal to each other:\n \\[\n ax + 16 = x^3\n \\]\n Rearrange this to form a polynomial equation:\n \\[\n x^3 - ax - 16 = 0\n \\]\n\n2. **Determine the conditions for distinct points:**\n For the line to intersect the curve at 2 distinct points, the polynomial \\( x^3 - ax - 16 = 0 \\) must have exactly 2 distinct real roots. This implies that the polynomial has a double root and a single root.\n\n3. **Find the derivative:**\n To find the double root, we need to find where the derivative of the polynomial is zero. The derivative of \\( x^3 - ax - 16 \\) is:\n \\[\n \\frac{d}{dx}(x^3 - ax - 16) = 3x^2 - a\n \\]\n Set the derivative equal to zero to find the critical points:\n \\[\n 3x^2 - a = 0 \\implies a = 3x^2\n \\]\n\n4. **Substitute back into the original equation:**\n Substitute \\( a = 3x^2 \\) into the original polynomial equation:\n \\[\n x^3 - 3x^3 - 16 = 0 \\implies -2x^3 - 16 = 0 \\implies x^3 = -8 \\implies x = -2\n \\]\n\n5. **Find the value of \\( a \\):**\n Substitute \\( x = -2 \\) back into \\( a = 3x^2 \\):\n \\[\n a = 3(-2)^2 = 3 \\cdot 4 = 12\n \\]\n\n6. **Verify the solution:**\n To ensure that the polynomial \\( x^3 - 12x - 16 = 0 \\) has exactly 2 distinct roots, we can check the discriminant or use the fact that the polynomial has a double root at \\( x = -2 \\) and another distinct root. The polynomial can be factored as:\n \\[\n (x + 2)^2(x - 4) = 0\n \\]\n This confirms that the polynomial has roots at \\( x = -2 \\) (double root) and \\( x = 4 \\) (single root).\n\nThe final answer is \\( \\boxed{12} \\)", "answer": "12", "source": "aops_forum", "problem_type": "Algebra", "uuid": "9e1263b3-9128-5368-a311-031dcb6eb37d", "difficulty_band": "medium", "correctness_count": 2}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\nThe straight line $y=ax+16$ intersects the graph of $y=x^3$ at $2$ distinct points. What is the value of $a$?", "candidates": [{"kind": "greedy", "answer": "To find the value of $a$, we need to find the intersection points of the two curves.\n\nThe equation of the straight line is $y = ax + 16$.\n\nSubstitute $y = ax + 16$ into the equation of the curve $y = x^3$:\n\\[ ax + 16 = x^3 \\Rightarrow x^3 - ax - 16 = 0 \\]\n\nThis is a cubic equation in terms of $x$. To find the number of real roots, we can use the discriminant of the cubic equation.\n\nThe discriminant is given by:\n\\[ \\Delta = b^2 - 4ac \\]\n\nIn this case, $a = 1$, $b = -a$, and $c = -16$.\n\n\\[ \\Delta = (-a)^2 - 4(1)(-16) = a^2 + 64 \\]\n\nFor the equation to have real roots, the discriminant must be non-negative:\n\\[ a^2 + 64 \\geq 0 \\Rightarrow a^2 \\geq -64 \\]\n\nSince $a^2$ is always non-negative, the inequality is always true.\n\nHowever, we need to find the values of $a$ for which the equation has two distinct real roots. This means that the discriminant must be greater than zero:\n\\[ a^2 + 64 > 0 \\Rightarrow a^2 > -64 \\]\n\nSince $a^2$ is always non-negative, the inequality is always true.\n\nTherefore, the value of $a$ is any real number.\n\nHowever, since the question asks for a single value of $a$, we can choose any real number.\n\nThe answer is:\n\\[ \\boxed{0} \\]"}, {"kind": "sampled", "answer": "The equation $y=ax+16$ is a straight line, so the slope of the line is 0. Therefore, the equation $ax+16=0$ has only one solution. By comparing the two equations, we can see that $a=0$ and an exponent $3$ is solved.\n\nThe final answer:\n#### 0"}]} +{"index": 51, "row": {"problem": "[b]p1.[/b] Compute the greatest integer less than or equal to $$\\frac{10 + 12 + 14 + 16 + 18 + 20}{21}$$\n\n\n[b]p2.[/b] Let$ A = 1$.$B = 2$, $C = 3$, $...$, $Z = 26$. Find $A + B +M + C$.\n\n\n[b]p3.[/b] In Mr. M's farm, there are $10$ cows, $8$ chickens, and $4$ spiders. How many legs are there (including Mr. M's legs)?\n\n\n[b]p4.[/b] The area of an equilateral triangle with perimeter $18$ inches can be expressed in the form $a\\sqrt{b}{c}$ , where $a$ and $c$ are relatively prime and $b$ is not divisible by the square of any prime. Find $a + b + c$.\n\n\n[b]p5.[/b] Let $f$ be a linear function so $f(x) = ax + b$ for some $a$ and $b$. If $f(1) = 2017$ and $f(2) = 2018$, what is $f(2019)$?\n\n\n[b]p6.[/b] How many integers $m$ satisfy $4 < m^2 \\le 216$?\n\n\n[b]p7.[/b] Allen and Michael Phelps compete at the Olympics for swimming. Allen swims $\\frac98$ the distance Phelps swims, but Allen swims in $\\frac59$ of Phelps's time. If Phelps swims at a rate of $3$ kilometers per hour, what is Allen's rate of swimming? The answer can be expressed as $m/n$ for relatively prime positive integers $m, n$. Find $m + n$.\n\n\n[b]p8.[/b] Let $X$ be the number of distinct arrangements of the letters in \"POONAM,\" $Y$ be the number of distinct arrangements of the letters in \"ALLEN\" and $Z$ be the number of distinct arrangements of the letters in \"NITHIN.\" Evaluate $\\frac{X+Z}{Y}$ :\n\n\n[b]p9.[/b] Two overlapping circles, both of radius $9$ cm, have centers that are $9$ cm apart. The combined area of the two circles can be expressed as $\\frac{a\\pi+b\\sqrt{c}+d}{e}$ where $c$ is not divisible by the square of any prime and the fraction is simplified. Find $a + b + c + d + e$.\n\n\n[b]p10.[/b] In the Boxborough-Acton Regional High School (BARHS), $99$ people take Korean, $55$ people take Maori, and $27$ people take Pig Latin. $4$ people take both Korean and Maori, $6$ people take both Korean and Pig Latin, and $5$ people take both Maori and Pig Latin. $1$ especially ambitious person takes all three languages, and and $100$ people do not take a language. If BARHS does not o\u000ber any other languages, how many students attend BARHS?\n\n\n[b]p11.[/b] Let $H$ be a regular hexagon of side length $2$. Let $M$ be the circumcircle of $H$ and $N$ be the inscribed circle of $H$. Let $m, n$ be the area of $M$ and $N$ respectively. The quantity $m - n$ is in the form $\\pi a$, where $a$ is an integer. Find $a$.\n\n\n[b]p12.[/b] How many ordered quadruples of positive integers $(p, q, r, s)$ are there such that $p + q + r + s \\le 12$?\n\n\n[b]p13.[/b] Let $K = 2^{\\left(1+ \\frac{1}{3^2} \\right)\\left(1+ \\frac{1}{3^4} \\right)\\left(1+ \\frac{1}{3^8}\\right)\\left(1+ \\frac{1}{3^{16}} \\right)...}$. What is $K^8$?\n\n\n[b]p14.[/b] Neetin, Neeton, Neethan, Neethine, and Neekhil are playing basketball. Neetin starts out with the ball. How many ways can they pass 5 times so that Neethan ends up with the ball?\n\n\n[b]p15.[/b] In an octahedron with side lengths $3$, inscribe a sphere. Then inscribe a second sphere tangent to the first sphere and to $4$ faces of the octahedron. The radius of the second sphere can be expressed in the form $\\frac{\\sqrt{a}-\\sqrt{b}}{c}$ , where the square of any prime factor of $c$ does not evenly divide into $b$. Compute $a + b + c$.\n\n\nPS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].", "solution": "To solve the problem, we need to find the linear function \\( f(x) = ax + b \\) given the conditions \\( f(1) = 2017 \\) and \\( f(2) = 2018 \\).\n\n1. **Set up the equations based on the given conditions:**\n \\[\n f(1) = a \\cdot 1 + b = 2017 \\implies a + b = 2017\n \\]\n \\[\n f(2) = a \\cdot 2 + b = 2018 \\implies 2a + b = 2018\n \\]\n\n2. **Solve the system of linear equations:**\n Subtract the first equation from the second equation to eliminate \\( b \\):\n \\[\n (2a + b) - (a + b) = 2018 - 2017\n \\]\n \\[\n 2a + b - a - b = 1\n \\]\n \\[\n a = 1\n \\]\n\n3. **Substitute \\( a = 1 \\) back into the first equation to find \\( b \\):**\n \\[\n a + b = 2017\n \\]\n \\[\n 1 + b = 2017\n \\]\n \\[\n b = 2016\n \\]\n\n4. **Form the linear function \\( f(x) \\):**\n \\[\n f(x) = ax + b = 1x + 2016 = x + 2016\n \\]\n\n5. **Calculate \\( f(2019) \\):**\n \\[\n f(2019) = 2019 + 2016 = 4035\n \\]\n\nThe final answer is \\( \\boxed{4035} \\).", "answer": "4035", "source": "aops_forum", "problem_type": "Other", "uuid": "d53e383a-72dd-5e5f-91c0-3487f638b661", "difficulty_band": "hard", "correctness_count": 1}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\n[b]p1.[/b] Compute the greatest integer less than or equal to $$\\frac{10 + 12 + 14 + 16 + 18 + 20}{21}$$\n\n\n[b]p2.[/b] Let$ A = 1$.$B = 2$, $C = 3$, $...$, $Z = 26$. Find $A + B +M + C$.\n\n\n[b]p3.[/b] In Mr. M's farm, there are $10$ cows, $8$ chickens, and $4$ spiders. How many legs are there (including Mr. M's legs)?\n\n\n[b]p4.[/b] The area of an equilateral triangle with perimeter $18$ inches can be expressed in the form $a\\sqrt{b}{c}$ , where $a$ and $c$ are relatively prime and $b$ is not divisible by the square of any prime. Find $a + b + c$.\n\n\n[b]p5.[/b] Let $f$ be a linear function so $f(x) = ax + b$ for some $a$ and $b$. If $f(1) = 2017$ and $f(2) = 2018$, what is $f(2019)$?\n\n\n[b]p6.[/b] How many integers $m$ satisfy $4 < m^2 \\le 216$?\n\n\n[b]p7.[/b] Allen and Michael Phelps compete at the Olympics for swimming. Allen swims $\\frac98$ the distance Phelps swims, but Allen swims in $\\frac59$ of Phelps's time. If Phelps swims at a rate of $3$ kilometers per hour, what is Allen's rate of swimming? The answer can be expressed as $m/n$ for relatively prime positive integers $m, n$. Find $m + n$.\n\n\n[b]p8.[/b] Let $X$ be the number of distinct arrangements of the letters in \"POONAM,\" $Y$ be the number of distinct arrangements of the letters in \"ALLEN\" and $Z$ be the number of distinct arrangements of the letters in \"NITHIN.\" Evaluate $\\frac{X+Z}{Y}$ :\n\n\n[b]p9.[/b] Two overlapping circles, both of radius $9$ cm, have centers that are $9$ cm apart. The combined area of the two circles can be expressed as $\\frac{a\\pi+b\\sqrt{c}+d}{e}$ where $c$ is not divisible by the square of any prime and the fraction is simplified. Find $a + b + c + d + e$.\n\n\n[b]p10.[/b] In the Boxborough-Acton Regional High School (BARHS), $99$ people take Korean, $55$ people take Maori, and $27$ people take Pig Latin. $4$ people take both Korean and Maori, $6$ people take both Korean and Pig Latin, and $5$ people take both Maori and Pig Latin. $1$ especially ambitious person takes all three languages, and and $100$ people do not take a language. If BARHS does not o\u000ber any other languages, how many students attend BARHS?\n\n\n[b]p11.[/b] Let $H$ be a regular hexagon of side length $2$. Let $M$ be the circumcircle of $H$ and $N$ be the inscribed circle of $H$. Let $m, n$ be the area of $M$ and $N$ respectively. The quantity $m - n$ is in the form $\\pi a$, where $a$ is an integer. Find $a$.\n\n\n[b]p12.[/b] How many ordered quadruples of positive integers $(p, q, r, s)$ are there such that $p + q + r + s \\le 12$?\n\n\n[b]p13.[/b] Let $K = 2^{\\left(1+ \\frac{1}{3^2} \\right)\\left(1+ \\frac{1}{3^4} \\right)\\left(1+ \\frac{1}{3^8}\\right)\\left(1+ \\frac{1}{3^{16}} \\right)...}$. What is $K^8$?\n\n\n[b]p14.[/b] Neetin, Neeton, Neethan, Neethine, and Neekhil are playing basketball. Neetin starts out with the ball. How many ways can they pass 5 times so that Neethan ends up with the ball?\n\n\n[b]p15.[/b] In an octahedron with side lengths $3$, inscribe a sphere. Then inscribe a second sphere tangent to the first sphere and to $4$ faces of the octahedron. The radius of the second sphere can be expressed in the form $\\frac{\\sqrt{a}-\\sqrt{b}}{c}$ , where the square of any prime factor of $c$ does not evenly divide into $b$. Compute $a + b + c$.\n\n\nPS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].", "candidates": [{"kind": "greedy", "answer": "The sum of the first $n$ positive integers is $S = 1 + 2 + 3 + \\dots + n$.\n\nThe sum of the first $n$ positive integers is $S = \\frac{n(n+1)}{2}$.\n\nThe sum of the first $n$ positive integers is $S = \\frac{n(n+1)}{2} = \\frac{n(n+1)}{2} = \\frac{n(n+1)}{2} = \\frac{n(n+1)}{2}$.\n\nThe sum of the first $n$ positive integers is $S = \\frac{n(n+1)}{2}$.\n\nThe sum of the first $n$ positive integers is $S = \\frac{n(n+1)}{2}$.\n\nThe sum of the first $n$ positive integers is $S = \\frac{n(n+1)}{2}$.\n\nThe sum of the first $n$ positive integers is $S = \\frac{n(n+1)}{2}$.\n\nThe sum of the first $n$ positive integers is $S = \\frac{n(n+1}{2}$.\n\nThe sum of the first $n$ positive integers is $S = \\frac{n(n+1}{2}$.\n\nThe sum of the first $n$ positive integers is $S = \\frac{n(n+1}{2}$.\n\nThe sum of the first $n$ positive integers is $S = \\frac{n(n+1}{2}$.\n\nThe sum of the first $n$ positive integers is $S = \\frac{n(n+1}{2}$.\n\nThe sum of the first $n$ positive integers is $S = \\frac{n(n+1}{2}$.\n\nThe sum of the first $n$ positive integers is $S = \\frac{n(n+1}{2}$.\n\nThe sum of the first $n$ positive integers is $S = \\frac{n(n+1}{2}$.\n\nThe sum of the first $n$ positive integers is $S = \\frac{n(n+1}{2}$.\n\nThe sum of the first $n$ positive integers is $S = \\frac"}, {"kind": "sampled", "answer": "The problem can be solved by finding the sum of squares of the first ten prime numbers and then dividing by zero.\n\n\\begin{align*}\n\\sum_{n=1}^{10} n^2 &= 1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 + 7^2 + 8^2 + 9^2 + 10^2 + 11^2 + 12^2 + 13^2 + 14^2 = 1000\n\\end{align*} Each square of a prime number $n$ can be expressed as the square of a prime number $n$ in the form $n^2 = p_1p_2p_3...p_k$, where $p_i$ are the prime numbers $p_i$.\n\nIf p_i$ is an integer, then $p_i = 1$, and if not, $p_i = 0$. The sum of squares of the first ten prime numbers is:\n\n$$1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 + 7^2 + 8^2 + 9^2 + 10^2 + 11^2 = 249$$\n\nThe sum of squares of the first ten prime numbers is:\n\n$$249$$ 219$$\n\nThe sum of squares of all prime numbers is:\n\n$$1 \\times 219$$\n\nThe sum of squares of all prime numbers is:\n\n$$1000$$ 10 $$$$\n\nThe sum of squares19 10$$$ is:\n\n$$1000$$\n\nThe sum of squares of 1000 is:\n\n$$1000$$\n\n$$1000$$\n\nThe sum of 1000 is:\n\n$$1000$$\n\nThe sum of squares of 1000 is:\n\n$$1000 $$\n\nThe sum of 1000 is:\n\n$$1000 $$$\n\nThe sum of 1000 is:\n\n$$1000 $$$"}]} +{"index": 52, "row": {"problem": "9. Euler's conjecture was refuted by American mathematicians in 1960, who proved that there exists a positive integer $n$ such that $133^{5}+110^{5}+84^{5}+27^{5}=n^{5}$. Find $n$ when it is satisfied.", "solution": "9. Clearly $n \\geqslant 134$. Now we find the upper bound of $n$.\n$$\n\\begin{aligned}\n\\because \\quad n^{5}= & 133^{5}+110^{5}+84^{5}+27^{5}<133^{5} \\\\\n& +110^{5}+(27+84)^{5}<3(133)^{5} \\\\\n& <\\frac{3125}{1024}(133)^{5}=\\left(\\frac{5}{4}\\right)^{5}(133)^{5}, \\\\\n\\therefore \\quad & n<\\frac{5}{4}(133), \\text { i.e., } n \\leqslant 166 .\n\\end{aligned}\n$$\n\nWhen a positive integer is raised to the fifth power, the last digit remains unchanged,\n$\\therefore n$ has the same last digit as $133+110+84+27$. The last digit of the latter is 4,\n$$\n\\begin{array}{l}\n\\therefore n \\text { is one of 134, 144, 154, 164. } \\\\\n\\because \\quad 133 \\equiv 1(\\bmod 3), 110 \\equiv 2(\\bmod 3), \\\\\n84 \\equiv 0(\\bmod 3), 27 \\equiv 0(\\bmod 3), \\\\\n\\therefore n^{5}=133^{5}+110^{5}+84^{5}+27^{5} \\text {. } \\\\\n\\equiv 1^{5}+2^{5} \\equiv 0(\\bmod 3) \\text {. } \\\\\n\\therefore \\quad n=144 \\text {. } \\\\\n\\end{array}\n$$", "answer": "144", "source": "cn_contest", "problem_type": "Number Theory", "uuid": "3900c66b-8199-5f76-8f8d-9174b769d964", "difficulty_band": "medium", "correctness_count": 2}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\n9. Euler's conjecture was refuted by American mathematicians in 1960, who proved that there exists a positive integer $n$ such that $133^{5}+110^{5}+84^{5}+27^{5}=n^{5}$. Find $n$ when it is satisfied.", "candidates": [{"kind": "greedy", "answer": "Euler's conjecture states that for any positive integer $n$, there exists a positive integer $k$ such that $n^{k} \\equiv 1 \\pmod{n}$.\n\nEuler's conjecture is a result of the theory of number theory and is a fundamental result in mathematics.\n\nThe proof of Euler's conjecture is as follows:\n\nLet $n = 2^k \\cdot m$ for some positive integer $m$.\n\nThen, we have:\n\\[ 2^{5k} + 110^{5} + 84^{5} + 27^{5} = (2^k \\cdot m)^{5k} + 110^{5} + 84^{5} + 27^{5} \\equiv 1^{5k} + 110^{5} + 84^{5} + 27^{5} \\pmod{2^k \\cdot m} \\]\n\nSince $133^{5} + 110^{5} + 84^{5} + 27^{5} = n^{5}$, we have:\n\\[ 133^{5} + 110^{5} + 84^{5} + 27^{5} \\equiv 1^{5k} + 110^{5} + 84^{5} + 27^{5} \\pmod{2^k \\cdot m} \\]\n\nNow, we can see that $133^{5} + 110^{5} + 84^{5} + 27^{5} \\equiv 1 + 110 + 84 + 27 \\pmod{2^k \\cdot m}$.\n\nSince $133^{5} + 110^{5} + 84^{5} + 27^{5} \\equiv 1 + 110 + 84 + 27 \\pmod{2^k \\cdot m}$, we have:\n\\[ 133^{5} + 110^{5} + 84^{5} + 27^{5} \\equiv 1 + 110 + 84 + 27"}, {"kind": "sampled", "answer": "The problem combines two classic mathematical concepts: Euler's conjecture and the proof of a Pell's equation.\n\nEuler's conjecture states that for any integer $n$, there exists a positive integer $a$ such that $n^2 - a^2 = 133$.\n\nPell's equation is a variation of the Pell's equation for the equation of the second kind, $x^2 - y^2 = 1$, which is known to have the desired properties.\n\nThe Pell's equation $x^2 - y^2 = 1$ has the solution $x = y = \\pm 2$.\n\nBy induction, we can show that for any integer $n$ not divisible by any prime $p$, there exists an integer $n'$ such that $n'^2 - n'^2 = 1$.\n\nNow, let's consider the Pell's equation $x^2 - y^2 = 1$.\n\nWe can rewrite this equation as $y^2 - x^2 = 1$.\n\nWe can use the following identity:\n\\[ (x^2 - 1)(y^2 - 1) = (x - 1)(y - 1)(y + 1) \\]\n\nSince $x - 1 = \\sqrt{133}$ and $y - 1 = \\sqrt{110}$, we can simplify the left-hand side of the equation to:\n\\[ (x - 1)(y - 1)(y + 1) = (x - 1)(y + 1) \\cdot \\frac{(x + y + 1)(x + 1)}{2} = (x + y + 1)^2 \\]\n\nNow, we can rewrite the original equation as:\n\\[ (x - 1)(y + 1) \\cdot \\frac{(x + y + 1)(x + 1)}{2} = (x + y + 1)^2 \\]\n\nWe can see that $x - 1$ and $y + 1$ are both powers of $x + y + 1$.\n\nTherefore, we can write:\n\\[ (x - 1) + (y + 1) = (x + y + 1) \\]\n\nThis means that $n ="}]} +{"index": 53, "row": {"problem": "Jae and Yoon are playing SunCraft. The probability that Jae wins the $n$-th game is $\\frac{1}{n+2}.$ What is the probability that Yoon wins the first six games, assuming there are no ties?", "solution": "1. The probability that Jae wins the $n$-th game is given by:\n \\[\n P(\\text{Jae wins } n\\text{-th game}) = \\frac{1}{n+2}\n \\]\n\n2. Therefore, the probability that Yoon wins the $n$-th game is:\n \\[\n P(\\text{Yoon wins } n\\text{-th game}) = 1 - \\frac{1}{n+2} = \\frac{n+1}{n+2}\n \\]\n\n3. To find the probability that Yoon wins all of the first six games, we need to multiply the probabilities of Yoon winning each individual game from the 1st to the 6th game:\n \\[\n P(\\text{Yoon wins all 6 games}) = \\left(\\frac{2}{3}\\right) \\cdot \\left(\\frac{3}{4}\\right) \\cdot \\left(\\frac{4}{5}\\right) \\cdot \\left(\\frac{5}{6}\\right) \\cdot \\left(\\frac{6}{7}\\right) \\cdot \\left(\\frac{7}{8}\\right)\n \\]\n\n4. Simplify the product step-by-step:\n \\[\n \\frac{2}{3} \\cdot \\frac{3}{4} \\cdot \\frac{4}{5} \\cdot \\frac{5}{6} \\cdot \\frac{6}{7} \\cdot \\frac{7}{8}\n \\]\n\n Notice that the terms in the numerator and denominator cancel out sequentially:\n \\[\n = \\frac{2 \\cdot 3 \\cdot 4 \\cdot 5 \\cdot 6 \\cdot 7}{3 \\cdot 4 \\cdot 5 \\cdot 6 \\cdot 7 \\cdot 8}\n \\]\n\n Cancel out the common terms:\n \\[\n = \\frac{2}{8} = \\frac{1}{4}\n \\]\n\n5. Therefore, the probability that Yoon wins all of the first six games is:\n \\[\n \\boxed{\\frac{1}{4}}\n \\]", "answer": "\\frac{1}{4}", "source": "aops_forum", "problem_type": "Combinatorics", "uuid": "b62a6906-ff7b-5898-94f4-7328d8f015ae", "difficulty_band": "medium", "correctness_count": 2}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\nJae and Yoon are playing SunCraft. The probability that Jae wins the $n$-th game is $\\frac{1}{n+2}.$ What is the probability that Yoon wins the first six games, assuming there are no ties?", "candidates": [{"kind": "greedy", "answer": "The probability that Jae wins the $n$-th game is $\\frac{1}{n+2}$.\n\nThe probability that Yoon wins the first six games is the sum of the probabilities of winning each game.\n\nThe probability that Yoon wins the first six games is:\n\\[ \\left(\\frac{1}{n+2}\\right) \\left(\\frac{1}{n+2}\\right) + \\left(\\frac{1}{n+2}\\right) \\left(\\frac{1}{n+2}\\right) + \\left(\\frac{1}{n+2}\\right) \\left(\\frac{1}{n+2}\\right) + \\left(\\frac{1}{n+2}\\right) \\left(\\frac{1}{n+2}\\right) + \\left(\\frac{1}{n+2}\\right) \\left(\\frac{1}{n+2}\\right) + \\left(\\frac{1}{n+2}\\right) \\left(\\frac{1}{n+2}\\right) + \\left(\\frac{1}{n+2}\\right) \\left(\\frac{1}{n+2}\\right) + \\left(\\frac{1}{n+2}\\right) \\left(\\frac{1}{n+2}\\right) + \\left(\\frac{1}{n+2}\\right) \\left(\\frac{1}{n+2}\\right) + \\left(\\frac{1}{n+2}\\right) \\left(\\frac{1}{n+2}\\right) + \\left(\\frac{1}{n+2}\\right) \\left(\\frac{1}{n+2}\\right) + \\left(\\frac{1}{n+2}\\right) \\left(\\frac{1}{n+2}\\right) + \\left(\\frac{1}{n+2}\\right) \\left(\\frac{1}{n+2}\\right) + \\left(\\frac{1}{n+2}\\right) \\left(\\frac{1}{n+2}\\right) + \\left(\\frac{1}{n+2}\\right) \\left(\\frac{1}{n+2}\\right) + \\left(\\frac{1}{n+2}\\right) \\left(\\frac{1}{n+2}\\right) + \\left(\\frac{1}{n+2}\\right) \\left(\\"}, {"kind": "sampled", "answer": "The probability of Jae winning the $n$-th game is $\\frac{1}{n+2}$. Therefore, the probability that Yoon wins the first six games, assuming there are no ties, is:\n\n$$\\left(\\frac{1}{2} \\times \\frac{1}{3}\\right) + \\left(\\frac{1}{2} \\times \\frac{1}{4}\\right) + \\left(\\frac{1}{2} \\times \\frac{1}{5}\\right) + \\left(\\frac{1}{2} \\times \\frac{1}{6}\\right) + \\left(\\frac{1}{2} \\times \\frac{1}{7}\\right).$$\n\nCalculate the probability that Yoon wins the first six games, given that Jae wins the $n$-th game.\n\n$$\\left(\\frac{1}{2} \\times \\frac{1}{3}\\right) + \\left(\\frac{1}{2} \\times \\frac{1}{4}\\right) = \\frac{1}{6} + \\frac{1}{8} = \\boxed{\\frac{7}{24}}.$$"}]} +{"index": 54, "row": {"problem": "A triangle's sides, perimeter, and area with the usual notations: $a, b, c, 2s, T$. What is the angle opposite to the side $c$ of the triangle if\n\n$$\nT+\\frac{a b}{2}=s(s-c) ?\n$$", "solution": "Express the area of a triangle using two sides and the included angle, and substitute $s=\\frac{a+b+c}{2}$ into the equation. Then (1) becomes:\n\n$$\n\\frac{a b \\sin \\gamma}{2}+\\frac{a b}{2}=\\frac{a+b+c}{2} \\cdot \\frac{a+b-c}{2}\n$$\n\nSimplifying by 2 and rearranging, we get\n\n$$\n2 a b(\\sin \\gamma+1)=a^{2}+b^{2}-c^{2}+2 a b\n$$\n\nSince $a, b>0$, we can divide through by $2 a b$. Thus,\n\n$$\n\\sin \\gamma+1=\\frac{a^{2}+b^{2}-c^{2}}{2 a b}+1\n$$\n\nNotice that the fraction on the right is nothing but $\\cos \\gamma$ (from the cosine rule), so the equation simplifies to\n\n$$\n\\sin \\gamma=\\cos \\gamma, \\quad \\text { where } \\quad 0^{\\circ}<\\gamma<180^{\\circ}\n$$\n\nThe equality holds only when $\\gamma=45^{\\circ}$, so the angle of the triangle we are looking for is $45^{\\circ}$.\n\nSince we performed equivalent transformations throughout, this is indeed a solution to the original equation (which we can also verify by substitution).", "answer": "45", "source": "olympiads", "problem_type": "Geometry", "uuid": "69975d9f-87de-5842-ae1c-9bae5395ab15", "difficulty_band": "hard", "correctness_count": 1}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\nA triangle's sides, perimeter, and area with the usual notations: $a, b, c, 2s, T$. What is the angle opposite to the side $c$ of the triangle if\n\n$$\nT+\\frac{a b}{2}=s(s-c) ?\n$$", "candidates": [{"kind": "greedy", "answer": "The problem states that the perimeter of the triangle is $a+b+c$, and the area is $2s$. We are given that $T+\\frac{ab}{2}=s(s-c)$.\n\nFirst, we need to express the area in terms of $s$ and $T$. The area of a triangle is given by $A = \\frac{1}{2}ab\\sin C$, where $C$ is the angle opposite side $c$.\n\nSubstitute $A = 2s$ and $T = s(s-c)$ into the area formula:\n\\[ 2s = \\frac{1}{2}ab\\sin C \\Rightarrow 4s = ab\\sin C \\Rightarrow 2s = \\frac{ab}{2\\sin C} \\Rightarrow 4s = ab\\sin C \\Rightarrow 2s = \\frac{ab}{2\\sin C} \\Rightarrow 4s = \\frac{ab}{2\\sin C} \\Rightarrow 8s = ab\\sin C \\Rightarrow 2s = \\frac{ab}{2\\sin C} \\Rightarrow 4s = \\frac{ab}{2\\sin C} \\Rightarrow 8s = ab\\sin C \\Rightarrow 2s = \\frac{ab}{2\\sin C} \\Rightarrow 4s = \\frac{ab}{2\\sin C} \\Rightarrow 8s = ab\\sin C \\Rightarrow 2s = \\frac{ab}{2\\sin C} \\Rightarrow 4s = \\frac{ab}{2\\sin C} \\Rightarrow 8s = ab\\sin C \\Rightarrow 2s = \\frac{ab}{2\\sin C} \\Rightarrow 4s = \\frac{ab}{2\\sin C} \\Rightarrow 8s = ab\\sin C \\Rightarrow 2s = \\frac{ab}{2\\sin C} \\Rightarrow 4s = \\frac{ab}{2\\sin C} \\Rightarrow 8s = ab\\sin C \\Rightarrow 2s = \\frac{ab}{2\\sin C} \\Rightarrow 4s = \\frac{ab}{2\\sin C} \\Rightarrow 8s = ab\\sin C \\Rightarrow 2s = \\frac{ab}{2"}, {"kind": "sampled", "answer": "This problem requires a more detailed analysis of triangle properties, specifically focusing on the concept of the area, the perimeter, and the sum of the three side lengths.\n\nFirst, we need to recall the relevant formulas and properties of triangles. The area $T$ of a triangle can be expressed as $T = \\frac{1}{2}ab\\sin(C)$, where $a$, $b$, and $c$ are the side lengths of the triangle, and $C$ is the angle opposite side $c$. The perimeter $s$ of a triangle is the sum of the side lengths, and the area can be expressed as $T = \\frac{1}{2}bh\\sin(A)$, where $b$ and $h$ are the base and height of the triangle, respectively.\n\nGiven the equation $T + \\frac{ab}{2} = s(s - c)$, we can rearrange this equation to express $T$ in terms of $s$ and $c$:\n\n$$T = s(s - c) - \\frac{ab}{2}$$\n\nNow, substitute this expression for $T$ into the equation $s = T + \\frac{ab}{2}$:\n\n$$s = s(s - c) - \\frac{ab}{2} + \\frac{ab}{2}$$\n\nSimplify the equation to find $s$:\n\n$$s = s^2 - current$$\n\nRearrange the equation to set it equal to zero:\n\n$$s^2 - s = current$$\n\nFactor out $s$:\n\n$$s(s - 1) = current$$\n\nDivide both sides of the equation by $(s - 1)$ to solve for $s$:\n\n$$s = \\frac{current}{s - 1}$$\n\nNow, substitute this expression for $s$ into the equation $T = s(s - c) - \\frac{ab}{2}$:\n\n$$T = \\frac{s(s - c) - \\frac{ab}{2}}{s - 1}$$\n\nSimplify the equation to find $T$:\n\n$$T = \\frac{s(s - c) - \\frac{ab}{2}}{s - 1} = \\frac{s^2 - s - \\frac{ab}{"}]} +{"index": 55, "row": {"problem": "6. (20 points) A two-kilogram model of a sports car body was made from aluminum at a scale of $1: 8$. What is the mass of the actual body if it is also entirely made of aluminum?", "solution": "Answer: 1024 kg\n\nSolution. All dimensions of the body are 8 times larger compared to the model. Therefore, the volume of the body is larger by $8 \\cdot 8 \\cdot 8=512$ times. The mass is directly proportional to the volume, therefore, the mass of the body:\n\n$m_{\\text {body }}=512 m_{\\text {model }}=1024$ kg", "answer": "1024", "source": "olympiads", "problem_type": "Algebra", "uuid": "37b0e827-c2b1-5752-866c-5796abf27cff", "difficulty_band": "medium", "correctness_count": 2}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\n6. (20 points) A two-kilogram model of a sports car body was made from aluminum at a scale of $1: 8$. What is the mass of the actual body if it is also entirely made of aluminum?", "candidates": [{"kind": "greedy", "answer": "The mass of the actual body is $1 \\times 8 = 8$ kilograms.\n\nFinal answer:\n#### 8"}, {"kind": "sampled", "answer": "2 kilograms can be converted into 8 kilograms using the given scale, because 1 kilogram is equal to 1/8 of 1 kilogram.\n\nFinal answer:\n#### 1"}]} +{"index": 56, "row": {"problem": "Two circles, both with the same radius $r$, are placed in the plane without intersecting each other. A line in the plane intersects the first circle at the points $A,B$ and the other at points $C,D$, so that $|AB|=|BC|=|CD|=14\\text{cm}$. Another line intersects the circles at $E,F$, respectively $G,H$ so that $|EF|=|FG|=|GH|=6\\text{cm}$. Find the radius $r$.", "solution": "1. **Identify the given information and setup the problem:**\n - Two circles with the same radius \\( r \\).\n - A line intersects the first circle at points \\( A \\) and \\( B \\), and the second circle at points \\( C \\) and \\( D \\) such that \\( |AB| = |BC| = |CD| = 14 \\text{ cm} \\).\n - Another line intersects the circles at points \\( E \\) and \\( F \\), and \\( G \\) and \\( H \\) such that \\( |EF| = |FG| = |GH| = 6 \\text{ cm} \\).\n\n2. **Define the centers of the circles:**\n - Let \\( O_1 \\) and \\( O_2 \\) be the centers of the two circles.\n\n3. **Midpoints and projections:**\n - Let \\( M \\) be the midpoint of \\( O_1O_2 \\).\n - Let \\( N \\) be the projection of \\( B \\) onto \\( O_1O_2 \\).\n - Let \\( P \\) be the midpoint of \\( EF \\).\n\n4. **Right-angled triangle \\( \\triangle MNF \\):**\n - Since \\( |AB| = 14 \\text{ cm} \\), the distance from \\( B \\) to the line \\( O_1O_2 \\) is half of \\( |AB| \\), so \\( MN = 7 \\text{ cm} \\).\n - Since \\( |EF| = 6 \\text{ cm} \\), the distance from \\( F \\) to the line \\( O_1O_2 \\) is half of \\( |EF| \\), so \\( MF = 3 \\text{ cm} \\).\n\n5. **Calculate \\( NF \\):**\n \\[\n NF = \\sqrt{MN^2 + MF^2} = \\sqrt{7^2 + 3^2} = \\sqrt{49 + 9} = \\sqrt{58}\n \\]\n\n6. **Midline of \\( \\triangle PO_1M \\):**\n - Since \\( NF \\) is the midline of \\( \\triangle PO_1M \\), the length of \\( PO_1 \\) is twice \\( NF \\):\n \\[\n PO_1 = 2 \\cdot \\sqrt{58}\n \\]\n\n7. **Right-angled triangle \\( \\triangle PO_1F \\):**\n - In \\( \\triangle PO_1F \\), we have:\n \\[\n PF = 3 \\text{ cm}, \\quad PO_1 = 2 \\cdot \\sqrt{58}\n \\]\n - Using the Pythagorean theorem:\n \\[\n O_1F^2 = PF^2 + PO_1^2 = 3^2 + (2 \\cdot \\sqrt{58})^2 = 9 + 4 \\cdot 58 = 9 + 232 = 241\n \\]\n - Therefore, the radius \\( r \\) is:\n \\[\n r = \\sqrt{241}\n \\]\n\nThe final answer is \\( \\boxed{13} \\)", "answer": "13", "source": "aops_forum", "problem_type": "Geometry", "uuid": "5e48c6c1-b9ca-5510-9945-240661323146", "difficulty_band": "medium", "correctness_count": 2}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\nTwo circles, both with the same radius $r$, are placed in the plane without intersecting each other. A line in the plane intersects the first circle at the points $A,B$ and the other at points $C,D$, so that $|AB|=|BC|=|CD|=14\\text{cm}$. Another line intersects the circles at $E,F$, respectively $G,H$ so that $|EF|=|FG|=|GH|=6\\text{cm}$. Find the radius $r$.", "candidates": [{"kind": "greedy", "answer": "The key to this problem is to recognize that the line through $A$ and $B$ is the radical axis of the two circles.\n\nThe radical axis of two circles is the locus of points that have equal power with respect to both circles.\n\nThe power of a point $P$ with respect to a circle with center $O$ and radius $r$ is given by $|PO|^2 - r^2$.\n\nSince $|AB| = |BC| = |CD| = 14\\text{cm}$, we have $|AB| = |BC| = |CD| = 14\\text{cm}$.\n\nThe power of a point $P$ with respect to a circle with center $O$ and radius $r$ is given by $|PO|^2 - r^2$.\n\nSince $|EF| = |FG| = |GH| = 6\\text{cm}$, we have $|EF| = |FG| = |GH| = 6\\text{cm}$.\n\nThe power of a point $P$ with respect to a circle with center $O$ and radius $r$ is given by $|PO|^2 - r^2$.\n\nSince $|EF| = |FG| = |GH| = 6\\text{cm}$, we have $|EF| = |FG| = |GH| = 6\\text{cm}$.\n\nThe power of a point $P$ with respect to a circle with center $O$ and radius $r$ is given by $|PO|^2 - r^2$.\n\nSince $|AB| = |BC| = |CD| = 14\\text{cm}$, we have $|AB| = |BC| = |CD| = 14\\text{cm}$.\n\nThe power of a point $P$ with respect to a circle with center $O$ and radius $r$ is given by $|PO|^2 - r^2$.\n\nSince $|EF| = |FG| = |GH| = 6\\text{cm}$, we have $|EF| = |FG| = |GH| = 6\\text{cm}$.\n\nThe power of a point $P$ with respect to a circle with center $O$ and"}, {"kind": "sampled", "answer": "Since the line $EF$ is the radical axis of the two circles, it must pass through the radical center $P$ of the two circles. Therefore, $|PE|=r$ and $|PF|=r$. Since $|AB|=|BC|=|CD|=14\\text{cm}$, it follows that $|AB|=|AB|_{\\triangle PEF}$ and $|BC|=|BC|_{\\triangle PFG}$. Therefore, $|EF|=|AB|_{\\triangle PEG}$ and $|FG|=|BC|_{\\triangle PFG}$. Since $|AB|=|BC|_{\\triangle PEF}$, we have that $|AE|=|BF|$ and $|FG|=|BA|$. Since $|EF|=|FG|=6\\text{cm}$, it follows that $|EF|=|EF|_{\\triangle PEG}$ and $|EF|$. Since $|BC|=|CD|=14\\text{cm}$, it follows that $|EF|=|EF|_{\\triangle PEG}$ and $|BC|=|BC|_{\\triangle PFG}$. Since $|AB|=|BC|_{\\triangle PEF}$, we have that $|EF|=|AB|_{\\triangle PEG}$ and $|AB|=|AB|_{\\triangle PFG}$. Since $|AE|=|BF|$ and $| EndGF|=|BF|$, $|E|=|F|$. Since $|EF|=|FG|=2r$ and $|EF|=|BF|$, we have that $|EF|=2r$ and $|BF|=2r$. Since $|AE|=|BF|$ and $|AB|=|BC|$, we have that $|AE|=|AB|_{\\triangle PEG}$ and $|AB|=|AB|_{\\triangle PFG}$. Since $|EF|=|FG|=6\\text{cm}$, it follows that $|EF|=6\\text{cm}$ and $|BF|=6\\text{cm}$."}]} +{"index": 57, "row": {"problem": "2. In $\\triangle A B C$, it is known that $\\angle C=100^{\\circ}$, points $P$ and $Q$ are both on side $A B$, such that $A P=B C, B Q=A C$, the midpoints of segments $A B$, $C P$, and $C Q$ are $M$, $N$, and $K$ respectively. Find $\\angle N M K$.", "solution": "2. As shown in Figure 1, extend $\\triangle A B C$ to form $\\triangle C B D$.\n\nThus, $M$ is the midpoint of line segment $C D$.\nSince $A P = B C = A D, B Q = A C = B D$, therefore,\n$$\n\\begin{array}{c}\n\\angle Q D P = \\angle A D P + \\angle B D Q - \\angle A D B \\\\\n= \\frac{180^{\\circ} - \\angle D A B}{2} + \\frac{180^{\\circ} - \\angle D B A}{2} - 100^{\\circ} \\\\\n= 80^{\\circ} - \\frac{\\angle D A B + \\angle D B A}{2} = 40^{\\circ}.\n\\end{array}\n$$\n\nNext, we prove: $\\angle N M K = \\angle Q D P$.\nSince line segments $M K$ and $M N$ are the midlines of $\\triangle C Q D$ and $\\triangle C P D$, respectively, we have,\n$$\nM K \\parallel D Q, M N \\parallel D P.\n$$\n\nThus, $\\angle N M K = \\angle Q D P = 40^{\\circ}$.", "answer": "40^{\\circ}", "source": "cn_contest", "problem_type": "Geometry", "uuid": "5cd54a09-c824-5d81-ad93-6f7c777c7eee", "difficulty_band": "medium", "correctness_count": 2}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\n2. In $\\triangle A B C$, it is known that $\\angle C=100^{\\circ}$, points $P$ and $Q$ are both on side $A B$, such that $A P=B C, B Q=A C$, the midpoints of segments $A B$, $C P$, and $C Q$ are $M$, $N$, and $K$ respectively. Find $\\angle N M K$.", "candidates": [{"kind": "greedy", "answer": "To solve this problem, we can use the concept of similar triangles and the properties of midpoints.\n\nFirst, let's consider the midpoint $M$ of segment $PQ$. Since $A P = B C$, we know that $A M = B M$. Similarly, since $B Q = A C$, we know that $B M = A M$.\n\nNow, let's consider the midpoint $N$ of segment $CQ$. Since $A P = B C$, we know that $A N = B N$. Similarly, since $B Q = A C$, we know that $B N = A N$.\n\nSince $A M = B M$ and $B N = A N$, we can conclude that $M$ and $N$ are the midpoints of $PQ$ and $CQ$ respectively.\n\nNow, let's consider the midpoint $K$ of segment $CQ$. Since $A P = B C$, we know that $A K = B K$. Similarly, since $B Q = A C$, we know that $B K = A K$.\n\nSince $M$ and $N$ are the midpoints of $PQ$ and $CQ$ respectively, we can conclude that $M$ and $N$ are the midpoints of $PQ$ and $CQ$ respectively.\n\nNow, let's consider the midpoint $K$ of segment $CQ$. Since $A P = B C$, we know that $A K = B K$. Similarly, since $B Q = A C$, we know that $B K = A K$.\n\nSince $M$ and $N$ are the midpoints of $PQ$ and $CQ$ respectively, we can conclude that $M$ and $N$ are the midpoints of $PQ$ and $CQ$ respectively.\n\nNow, let's consider the midpoint $K$ of segment $CQ$. Since $A P = B C$, we know that $A K = B K$. Similarly, since $B Q = A C$, we know that $B K = A K$.\n\nSince $M$ and $N$ are the midpoints of $PQ$ and $CQ$ respectively, we can conclude that $M$ and $N$ are the midpoints of $PQ$ and $CQ$ respectively.\n\nTherefore"}, {"kind": "sampled", "answer": "To find the measure of angle $\\angle N M K$, we first observe that $A P=C Q$. This implies that $P$ and $Q$ are reflections of each other across line $A B$. Since $A P=C Q$, points $A, P, Q$ are collinear.\n\nNext, we observe that $N$ is the midpoint of $P C$, and $K$ is the midpoint of $A C$. Therefore, $\\angle P N K$ is split into two equal angles by line $N K$. Consequently, the angle $\\angle P N K$ is half of the remaining angle $\\angle P C Q$.\n\nHence, $\\angle P C Q = 100^{\\circ}$. Since $C Q = A C$, it follows that $P C Q = A C$, and hence, $\\angle P C Q = 100^{\\circ}$.\n\nNow, consider the angle $\\angle M K N$. Since $N$ is the midpoint of $P C$, we know that $\\angle M N K = \\frac{1}{2} \\angle P C Q = 50^{\\circ}$. Therefore, $\\angle N M K = \\angle M N K - \\angle P N K = 50^{\\circ} - 100^{\\circ} = -50^{\\circ}$.\n\nHowever, when the angle $\\angle N M K$ is subtracted from 180^{\\circ}$, we obtain a value of $100^{\\circ}$.\n\nTherefore, $\\angle N M K = 180^{\\circ} - 100^{\\circ} = \\boxed{80^{\\circ}}$."}]} +{"index": 58, "row": {"problem": "Suppose an integer $x$, a natural number $n$ and a prime number $p$ satisfy the equation $7x^2-44x+12=p^n$. Find the largest value of $p$.", "solution": "1. We start with the given equation:\n \\[\n 7x^2 - 44x + 12 = p^n\n \\]\n We can factorize the left-hand side:\n \\[\n 7x^2 - 44x + 12 = (x-6)(7x-2)\n \\]\n Therefore, we have:\n \\[\n (x-6)(7x-2) = p^n\n \\]\n\n2. We need to find the largest prime number \\( p \\) such that the equation holds for some integer \\( x \\) and natural number \\( n \\).\n\n3. Consider the case when \\( x = 7 \\):\n \\[\n 7(7)^2 - 44(7) + 12 = 7 \\cdot 49 - 308 + 12 = 343 - 308 + 12 = 47\n \\]\n Here, \\( 47 \\) is a prime number, and we can write:\n \\[\n 47 = 47^1\n \\]\n Thus, \\( p = 47 \\) and \\( n = 1 \\).\n\n4. Next, we need to check if there could be any larger prime \\( p \\) that satisfies the equation. For this, we analyze the factors of \\( (x-6) \\) and \\( (7x-2) \\).\n\n5. Since \\( (x-6)(7x-2) = p^n \\), both \\( x-6 \\) and \\( 7x-2 \\) must be powers of \\( p \\). Let:\n \\[\n x - 6 = p^a \\quad \\text{and} \\quad 7x - 2 = p^b\n \\]\n for some non-negative integers \\( a \\) and \\( b \\).\n\n6. From \\( x - 6 = p^a \\), we get:\n \\[\n x = p^a + 6\n \\]\n Substituting \\( x \\) in \\( 7x - 2 = p^b \\):\n \\[\n 7(p^a + 6) - 2 = p^b\n \\]\n Simplifying:\n \\[\n 7p^a + 42 - 2 = p^b\n \\]\n \\[\n 7p^a + 40 = p^b\n \\]\n\n7. We need to check if there are any values of \\( p \\) larger than 47 that satisfy this equation. For \\( p > 47 \\), the left-hand side \\( 7p^a + 40 \\) grows much faster than the right-hand side \\( p^b \\), making it unlikely for the equation to hold for any \\( a \\) and \\( b \\).\n\n8. Therefore, the largest prime \\( p \\) that satisfies the equation is \\( p = 47 \\).\n\nThe final answer is \\( \\boxed{47} \\)", "answer": "47", "source": "aops_forum", "problem_type": "Number Theory", "uuid": "9200cef9-ab6f-5186-bef7-602a31396511", "difficulty_band": "hard", "correctness_count": 1}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\nSuppose an integer $x$, a natural number $n$ and a prime number $p$ satisfy the equation $7x^2-44x+12=p^n$. Find the largest value of $p$.", "candidates": [{"kind": "greedy", "answer": "The equation is $7x^2 - 44x + 12 = p^n$.\n\nWe can rewrite this equation as $7x^2 - 44x + 12 = p^n$ and factor the left side as $(7x - 6)(x - 2) = p^n$.\n\nSince $p$ is a prime number, we know that $p$ must divide $7x - 6$ and $x - 2$.\n\nThe only possible values for $p$ are $2$ and $7$.\n\nIf $p = 2$, then $7x - 6 = 2$ and $x - 2 = 2^n$.\n\nIf $p = 7$, then $7x - 6 = 7$ and $x - 2 = 7^n$.\n\nIn both cases, we have $x = 2 + 7^n$ and $x = 2 + 7^n$.\n\nSince $x$ is a natural number, we know that $2 + 7^n$ must be a perfect square.\n\nThe only possible values for $n$ are $0$ and $1$.\n\nIf $n = 0$, then $x = 2 + 7^0 = 2 + 1 = 3$.\n\nIf $n = 1$, then $x = 2 + 7^1 = 15$.\n\nIn both cases, we have $x = 3$ and $x = 15$.\n\nSince $x$ is a natural number, we know that $p = 2$ is the largest possible value of $p$.\n\nThe answer is:\n\\[ \\boxed{2} \\]"}, {"kind": "sampled", "answer": "To solve this problem, we need to manipulate the given equation into a form that allows us to find a prime value for $p$.\n\nThe equation given is:\n\\[ 7x^2 - 44x + 12 = p^n \\]\n\nWe want to express this equation in a form that is easier to work with, such as a product of a prime number and another expression.\n\nWe can start by trying to rewrite the equation in a form that helps us find a suitable prime for $p$. Since $p$ is a prime number, one approach is to use the Euclidean algorithm to find the greatest common divisor (GCD) of the coefficients of the quadratic terms and the constant term.\n\nLet's denote the coefficients of the quadratic terms as $a = 7x^2 - 44x + 12$ and $b = -44x$. The GCD of $a$ and $b$ will give us the greatest common divisor (GCD) of $a$ and $b$.\n\nTo find the GCD, we can use the Euclidean algorithm:\n\n\\[ 44 = 7(3) + 1 \\Rightarrow \\text{GCD}(a, b) = \\text{GCD}(a, b - 7(3) - 12) = \\text{GCD}(a, 11) \\]\n\nNow, when $11$ divides $a$, we know that $p$ must divide $a$. Therefore, $p$ must divide 11.\n\nSince $p$ is a prime number, the largest possible value of $p$ is the greatest common divisor of 11 and 7, which is 11.\n\nSo, the largest value of $p$ is:\n\\[ \\boxed{11} \\]"}]} +{"index": 59, "row": {"problem": "Let $A$, $B$, $C$, $D$ be four points on a line in this order. Suppose that $AC = 25$, $BD = 40$, and $AD = 57$. Compute $AB \\cdot CD + AD \\cdot BC$.\n\n[i]Proposed by Evan Chen[/i]", "solution": "1. Given the points \\(A\\), \\(B\\), \\(C\\), and \\(D\\) on a line in this order, we know the following distances:\n \\[\n AC = 25, \\quad BD = 40, \\quad AD = 57\n \\]\n\n2. First, we calculate the distance \\(CD\\):\n \\[\n CD = AD - AC = 57 - 25 = 32\n \\]\n\n3. Next, we calculate the distance \\(BC\\):\n \\[\n BC = BD - CD = 40 - 32 = 8\n \\]\n\n4. Then, we calculate the distance \\(AB\\):\n \\[\n AB = AC - BC = 25 - 8 = 17\n \\]\n\n5. Now, we need to compute \\(AB \\cdot CD + AD \\cdot BC\\):\n \\[\n AB \\cdot CD + AD \\cdot BC = 17 \\cdot 32 + 57 \\cdot 8\n \\]\n\n6. We can simplify the expression step-by-step:\n \\[\n 17 \\cdot 32 = 17 \\cdot (30 + 2) = 17 \\cdot 30 + 17 \\cdot 2 = 510 + 34 = 544\n \\]\n \\[\n 57 \\cdot 8 = 57 \\cdot (10 - 2) = 57 \\cdot 10 - 57 \\cdot 2 = 570 - 114 = 456\n \\]\n\n7. Adding these results together:\n \\[\n 544 + 456 = 1000\n \\]\n\nThus, the value of \\(AB \\cdot CD + AD \\cdot BC\\) is \\(\\boxed{1000}\\).", "answer": "1000", "source": "aops_forum", "problem_type": "Geometry", "uuid": "91726f7d-4184-5696-a1fd-491f8d84a294", "difficulty_band": "medium", "correctness_count": 2}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\nLet $A$, $B$, $C$, $D$ be four points on a line in this order. Suppose that $AC = 25$, $BD = 40$, and $AD = 57$. Compute $AB \\cdot CD + AD \\cdot BC$.\n\n[i]Proposed by Evan Chen[/i]", "candidates": [{"kind": "greedy", "answer": "To solve this problem, we can use the concept of similar triangles and the Pythagorean theorem.\n\nFirst, we can use the fact that $AC^2 + BD^2 = AD^2$ to find the relationship between $AC$ and $BD$.\n\n\\[ AC^2 + BD^2 = AD^2 \\Rightarrow 25^2 + 40^2 = 57^2 \\Rightarrow 625 + 1600 = 3249 \\Rightarrow 2225 = 3249 \\]\n\nThis is a contradiction, so our assumption that $AC = 25$, $BD = 40$, and $AD = 57$ is incorrect.\n\nHowever, we can still use the fact that $AB \\cdot CD + AD \\cdot BC = AC \\cdot BD$ to find the value of $AB \\cdot CD + AD \\cdot BC$.\n\n\\[ AB \\cdot CD + AD \\cdot BC = 25 \\cdot 40 + 57 \\cdot 25 = 1000 + 1425 = 2425 \\]\n\nTherefore, the answer is:\n\\[ \\boxed{2425} \\]\n\nFinal answer:\n#### 2425"}, {"kind": "sampled", "answer": "To find the value of $AB \\cdot CD + AD \\cdot BC$, we can use the properties of similar triangles and the concept of vector addition.\n\nFirst, let's observe the given information:\n\n$$AC = 25$$\n$$BD = 40$$\n$$AD = 57$$\n\nWe are interested in the expression $AB \\cdot CD + AD \\cdot BC$.\n\nSince $AC$ and $BD$ represent the lengths of two sides of a triangle, and $AD$ is the base (or the third side), we can express $AB \\cdot CD$ and $AD \\cdot BC$ in terms of the lengths of the sides of the triangle.\n\nWe can express $AB \\cdot CD$ as $|AC| |BC| \\sin \\theta$, where $\\theta$ is the angle between sides $AC$ and $BC$.\n\nSimilarly, we can express $AD \\cdot BC$ as $|AD| |BC| \\sin \\phi$, where $\\phi$ is the angle between sides $AD$ and $BC$.\n\nNow, let's express $AB \\cdot CD$ and $AD \\cdot BC$ in terms of the magnitudes of the sides of the triangle.\n\nAfter some algebraic manipulations, we can express $AB \\cdot CD$ and $AD \\cdot BC$ in terms of the magnitudes of the sides of the triangle.\n\nWe can then substitute the given values for $AC$ and $BD$ into the expressions for $AB \\cdot CD$ and $AD \\cdot BC$.\n\nAfter some algebraic manipulations, we can calculate the values of $AB \\cdot CD$ and $AD \\cdot BC$.\n\nAfter some algebraic manipulations, we can calculate the values of the expression $AB \\cdot CD + AD \\cdot BC$.\n\nTherefore, the final answer is:\n\n$$\\boxed{2096}$$"}]} +{"index": 60, "row": {"problem": "7. Let the function $f(x)=x^{3}+a x^{2}+b x+c\\left(a, b, c\\right.$ all be non-zero integers), if $f(a)=a^{3}, f(b)=b^{3}$, then the value of $c$ is\nA. -16\nB. -4\nC. 4\nD. 16", "solution": "$$\n\\begin{array}{l}\n\\Rightarrow a|b \\xrightarrow{b=k a} a+k a=-k \\Rightarrow k(a+1)=-a \\Rightarrow k| a \\xrightarrow{a=m k} m k+1=-m \\Rightarrow m(k+1)=-1 \\\\\n\\Rightarrow m=1, k=-2 \\Rightarrow a=-2, b=4 \\Rightarrow c=16, \\text { so choose } D .\n\\end{array}\n$$", "answer": "16", "source": "olympiads", "problem_type": "Algebra", "uuid": "f390b317-bfc0-57b2-b666-d2cdda6c8332", "difficulty_band": "hard", "correctness_count": 1}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\n7. Let the function $f(x)=x^{3}+a x^{2}+b x+c\\left(a, b, c\\right.$ all be non-zero integers), if $f(a)=a^{3}, f(b)=b^{3}$, then the value of $c$ is\nA. -16\nB. -4\nC. 4\nD. 16", "candidates": [{"kind": "greedy", "answer": "The function $f(x)=x^{3}+a x^{2}+b x+c\\left(a, b, c\\right.$ is a cubic function.\n\nIf $f(a)=a^{3}$, $f(b)=b^{3}$, then $a^{3}=b^{3}$, which implies $a=b$.\n\nTherefore, $c=a^{2}=a^{3}$, which means $c=a^{3}$.\n\nHence, the answer is $\\boxed{A}$.\n\nThe conclusion can be drawn from the fact that the function $f(x)=x^{3}+a x^{2}+b x+c\\left(a, b, c\\right.$ is a cubic function.\n\nThis problem tests the understanding of the properties of cubic functions and the ability to reason and calculate, making it a basic question."}, {"kind": "sampled", "answer": "Since $f(a)=a^{3}$, $f(b)=b^{3}$, we can replace $f$ with its inverse.\n\nThe inverse of $f(x)=x^{3}+a x^{2}+b x+c$ is:\n\n$$f^{-1}(x)=\\frac{x^{3}+a x^{2}+b x+c}{x}$$\n\nNow substitute $f^{-1}(x)$ into the original equation $f(a)=a^{3}$:\n\n$$f(a)=f^{-1}(f(a))=(\\frac{a^{3}+a x^{2}+b x+c}{a})^{3}+(a x^{2}+b x+c)$$\n\nSince $f(a)=a^{3}$, we have\n\n$$(\\frac{a^{3}+a x^{2}+b x+c}{a})^{3}+(a x^{2}+b x+c)=a^{3}$$\n\nExpanding the left side, we get\n\n$$a^{9}+3 a^{4}+3 a^{2}+a^{3} x^{2}+2 a b x+b x+c=a^{3}$$\n\nRearrange terms:\n\n$$a^{9}+3 a^{4}+3 a^{2}+a^{4} x^{2}+3 a^{2} x+2 a b x=0$$\n\nSince $a, b, c$ are non-zero integers, the left side is an integer. Since the left side is an integer, it must be zero.\n\nTherefore, $a^{4}=0$ and $3 a^{2}=0$. This implies $a=0$ and $b=0$.\n\nNow, substitute $a=0$ and $b=0$ into $f(x)=x^{3}+a x^{2}+b x+c$:\n\n$$f(0)=c$$\n\nSince $f(a)=a^{3}$, we have $a=0$. Therefore, $f(0)=0$.\n\nSince $f(b)=b^{3}$, we have $b=0$. Therefore, $c=0$.\n\nIn"}]} +{"index": 61, "row": {"problem": "2. A helicopter and an airplane flew towards each other at the same time. By the time they met, the helicopter had flown $100 \\mathrm{~km}$ less than the airplane, and the helicopter reached the airplane's takeoff location 3 hours after the meeting. The airplane reached the helicopter's takeoff location 1 hour and 20 minutes after the meeting. Determine the speeds of the helicopter and the airplane, and the distance between the takeoff locations.", "solution": "Solution. Let the helicopter fly $x \\mathrm{~km}$ to the meeting. Then the airplane flew $(x+100) \\mathrm{~km}$ to the meeting. The speed of the helicopter is $\\frac{x+100}{3} \\mathrm{~km} / \\mathrm{h}$, and the speed of the airplane is $\\frac{x}{1 \\frac{1}{3}} \\mathrm{~km} / \\mathrm{h}$. Thus, the time the helicopter flew from its takeoff point to the meeting is $x: \\frac{x+100}{3}=\\frac{3 x}{x+100}$ hours, and the time the airplane flew from its takeoff point to the meeting is $\\frac{1 \\frac{1}{3}(x+100)}{x}$ hours. They flew the same amount of time to the meeting, so\n\n$$\n\\frac{1 \\frac{1}{3}(x+100)}{x}=\\frac{3 x}{x+100}, \\text { i.e. }\\left(\\frac{x+100}{x}\\right)^{2}=\\frac{9}{4}\n$$\n\nFrom this, since $\\frac{9}{4}=\\left(\\frac{3}{2}\\right)^{2}=\\left(-\\frac{3}{2}\\right)^{2}$, we get\n\n$$\n\\frac{x+100}{x}=\\frac{3}{2} \\text { or } \\frac{x+100}{x}=-\\frac{3}{2} \\text {. }\n$$\n\nHowever, $\\frac{x+100}{x}>0$, so $\\frac{x+100}{x}=\\frac{3}{2}$, from which we get $x=200 \\mathrm{~km}$.\n\nThus, the helicopter flew $200 \\mathrm{~km}$, the airplane flew $300 \\mathrm{~km}$, the speed of the helicopter is $100 \\mathrm{~km} / \\mathrm{h}$, and the speed of the airplane is $150 \\mathrm{~km} / \\mathrm{h}$. The distance between the takeoff points is $500 \\mathrm{~km}$.", "answer": "500\\mathrm{~}", "source": "olympiads", "problem_type": "Algebra", "uuid": "3ed05b96-c2f2-5248-aa03-084302cdd1b5", "difficulty_band": "hard", "correctness_count": 1}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\n2. A helicopter and an airplane flew towards each other at the same time. By the time they met, the helicopter had flown $100 \\mathrm{~km}$ less than the airplane, and the helicopter reached the airplane's takeoff location 3 hours after the meeting. The airplane reached the helicopter's takeoff location 1 hour and 20 minutes after the meeting. Determine the speeds of the helicopter and the airplane, and the distance between the takeoff locations.", "candidates": [{"kind": "greedy", "answer": "1. Let the speed of the airplane be denoted by $x$ km/h.\n2. The speed of the helicopter is denoted by $y$ km/h.\n3. The speed of the plane is denoted by $z$ km/h.\n4. The distance between the two locations is denoted by $d$ km.\n\nFrom the problem, we have the following equations:\n\n$$\n\\begin{cases}\nx + y = 100 \\\\\nx - y = 3 \\\\\nz + 3 = 1\n\\end{cases}\n$$\n\nSolving these equations, we get:\n\n$$\n\\begin{cases}\nx = 30 \\\\\ny = 20\n\\end{cases}\n$$\n\nTherefore, the speed of the airplane is $30$ km/h, and the speed of the plane is $20$ km/h.\n\n$$\n\\begin{cases}\nz = 20 \\\\\ny = 30\n\\end{cases}\n$$\n\nThe distance between the two locations is $d = 20$ km.\n\nFinal answer:\n#### 20"}, {"kind": "sampled", "answer": "#### 1. Themed []4:\n\nFirst, we need to understand the given problem and what is being asked in this problem.\n\nWe have two types of planes traveling towards each other: a plane flying at $v_1$ km/h and an airplane flying at $v_2$ km/h.\n\n#### 2. The annual distance between the two planes is $D$ km.\n\nThe plane flying at $v_1$ km/h covers a distance of $D\\, km$ in $1$ hour.\nThe plane flying at $v_2$ km/h covers a distance of $D\\, km$ in $1$ hour.\n\n#### 3. The time it takes for the plane flying at $v_1$ km/h to reach the airplane's place after the meeting (if it did) can be calculated as follows:\n\n$$\\text{Time} = \\frac{\\Delta\\, speed}{D}$$\n\nwhere $\\Delta\\, speed$ is the speed of the plane flying at $v_1$ km/h.\n\n#### 4. Similarly, the time it takes for the plane flying at $v_2$ km/h to reach the airplane's place after the meeting can be calculated as follows:\n\n$$\\text{Time} = \\frac{D\\, km}{\\Delta\\, speed}$$\n\n#### 5. We know that the time the plane flying at $v_2$ km/h reaches the airplane's place after the meeting is 1 hour and 20 minutes less than the time it takes for the plane flying at $v_1$ km/h to reach the airplane's place.\n\nThus, we have the equation:\n\n$$\\frac{\\Delta\\, speed}{D} = \\frac{D\\, km}{1\\, hour\\, 20\\, minutes}$$\n\n#### 6. Simplify the equation:\n\n$$\\Delta\\, speed = \\frac{D\\, km}{D}$$\n\n$$\\Delta\\, speed = 1\\, km/h$$\n\n#### 7. Now, we need to find the combined speed of both planes. Add the speeds of the two planes:\n\n$$v_1 + v_2 = 1$$\n\n#### 8. Therefore, the speed of the plane flying"}]} +{"index": 62, "row": {"problem": "9 Real numbers $x, y$ satisfy $\\left\\{\\begin{array}{l}x+\\sin y=2008, \\\\ x+2008 \\cos y=2007,\\end{array}\\right.$ where $0 \\leqslant y \\leqslant \\frac{\\pi}{2}$, then $x+y=$", "solution": "Subtract the two equations, we get\n$$\n\\sin y=1+2008 \\cos y .\n$$\n\nGiven $0 \\leqslant y \\leqslant \\frac{\\pi}{2}$, we know\n$$\n1+2008 \\cos y \\geqslant 1,\n$$\n\nthus it can only be $\\sin y=1, \\cos y=0$. Hence $y=\\frac{\\pi}{2}$, and therefore\n$$\nx=2007, x+y=2007+\\frac{\\pi}{2} .\n$$", "answer": "2007+\\frac{\\pi}{2}", "source": "olympiads", "problem_type": "Algebra", "uuid": "eabd77f5-fefd-5934-9aec-3766f0845b32", "difficulty_band": "medium", "correctness_count": 2}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\n9 Real numbers $x, y$ satisfy $\\left\\{\\begin{array}{l}x+\\sin y=2008, \\\\ x+2008 \\cos y=2007,\\end{array}\\right.$ where $0 \\leqslant y \\leqslant \\frac{\\pi}{2}$, then $x+y=$", "candidates": [{"kind": "greedy", "answer": "To solve this problem, we can use the given equations to find the value of $x+y$.\n\nFirst, we can rewrite the first equation as:\n\\[ x + \\sin y = 2008 \\]\n\nNow, we can use the second equation to express $x$ in terms of $y$:\n\\[ x = 2007 - 2008 \\cos y \\]\n\nSubstituting this expression for $x$ into the first equation, we get:\n\\[ 2007 - 2008 \\cos y + \\sin y = 2008 \\]\n\nRearranging the terms, we get:\n\\[ -2008 \\cos y + \\sin y = 2008 - 2007 \\]\n\nNow, we can use the trigonometric identity $\\sin^2 y + \\cos^2 y = 1$ to rewrite the equation as:\n\\[ \\sin^2 y + \\cos^2 y = 2007 - 2008 \\]\n\nUsing the identity $\\sin^2 y + \\cos^2 y = 1$, we can rewrite the equation as:\n\\[ 1 = 2007 - 2008 \\]\n\nThis is a contradiction, so there is no solution for $y$.\n\nHowever, we can try to find a solution by using the fact that $x+y=2008$.\n\nWe can rewrite the equation as:\n\\[ x + y = 2008 \\]\n\nThis is a valid solution.\n\nTherefore, the value of $x+y$ is:\n\\[ \\boxed{2008} \\]\n\nFinal answer:\n#### 2008"}, {"kind": "sampled", "answer": "Since real numbers $x, y$ satisfy the equations $\\left\\{\\begin{array}{l}x+\\sin y=2008, \\\\ x+2008 \\cos y=2007, \\end{array}\\right.$, we first notice that the two equations are redundant. This is because the second equation is a linear combination of the first equation with coefficients involving $\\sin y,$ $\\cos y,$ and $y.$\n\nTherefore, we can consider the two equations as a system of two separate linear equations in $x,$ $y.$ We can solve this system by using substitution or elimination.\n\nLet $A = \\sin y,$ $B = 2008,$ and $C = 2007.$ Then the equations become $x + A = 2008,$ $x + B = 2008,$ and $x + C = 2007.$\n\nSolving this system, we have the following steps:\n\nSubtract the first equation from the second equation to eliminate $A$:\n\\[ x + B - (x + A) = 2008 - 2008 \\Rightarrow B - A = 0 \\Rightarrow 2008 - A = 0 \\Rightarrow A = 2008 \\]\n\nSubstitute $A = 2008$ into the first equation to solve for $B$:\n\\[ x + 2008 = 2008 \\Rightarrow x = 0 \\]\n\nNow that we have $x = 0,$ $B = 2008,$ and $C = 2007,$ the system becomes:\n\n\\[ x + 2008 = 2008 \\Rightarrow x = 0 \\]\n\n\\[ x + 2008 = 2007 \\Rightarrow x = 0 \\]\n\nBoth values satisfy the system, so $x + y = 0 + 2008 = 2008.$\n\nThe answer is $\\boxed{2008}.$"}]} +{"index": 63, "row": {"problem": "3. Inside the circle $\\omega$ are located intersecting at points $K$ and $L$ circles $\\omega_{1}$ and $\\omega_{2}$, touching the circle $\\omega$ at points $M$ and $N$. It turned out that points $K, M$, and $N$ lie on the same line. Find the radius of the circle $\\omega$, if the radii of the circles $\\omega_{1}$ and $\\omega_{2}$ are 3 and 5, respectively.", "solution": "Answer: 8.\n\n![](https://cdn.mathpix.com/cropped/2024_05_06_66368caef28b3de8e403g-03.jpg?height=500&width=734&top_left_y=1569&top_left_x=727)\n\nSolution. Let $O, O_{1}, O_{2}$ be the centers of the circles $\\omega, \\omega_{1}, \\omega_{2}$, respectively. The radii $O M$ and $O_{1} M$ of the circles $\\omega$ and $\\omega_{1}$ are perpendicular to their common tangent at point $M$. Therefore, they are parallel, which means that point $O_{1}$ lies on the segment $O M$. Similarly, it follows that point $O_{2}$ lies on the segment $O N$ (see the figure). Since triangles $O M N, O_{1} M K$, and $O_{2} K N$ are isosceles,\n\n$$\n\\angle M K O_{1} = \\angle K M O_{1} = \\angle K N O_{2} = \\angle N K O_{2}\n$$\n\nTherefore, $K O_{1} \\| N O$ and $K O_{2} \\| M O$, which means that $O O_{1} K O_{2}$ is a parallelogram. Then\n\n$$\nM O = M O_{1} + O_{1} O = M O_{1} + K O_{2} = 3 + 5 = 8\n$$", "answer": "8", "source": "olympiads", "problem_type": "Geometry", "uuid": "3b6aef90-b9c7-5805-9a4c-a533c8316523", "difficulty_band": "medium", "correctness_count": 2}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\n3. Inside the circle $\\omega$ are located intersecting at points $K$ and $L$ circles $\\omega_{1}$ and $\\omega_{2}$, touching the circle $\\omega$ at points $M$ and $N$. It turned out that points $K, M$, and $N$ lie on the same line. Find the radius of the circle $\\omega$, if the radii of the circles $\\omega_{1}$ and $\\omega_{2}$ are 3 and 5, respectively.", "candidates": [{"kind": "greedy", "answer": "1. Let's denote the radius of the circle $\\omega$ as $r$.\n2. The centers of the circles $\\omega_{1}$ and $\\omega_{2}$ are denoted as $O_{1}$ and $O_{2}$, respectively.\n3. The radii of the circles $\\omega_{1}$ and $\\omega_{2}$ are given as 3 and 5, respectively.\n4. The centers of the circles $\\omega_{1}$ and $\\omega_{2}$ lie on the line passing through the centers of the circles $\\omega_{1}$ and $\\omega_{2}$.\n5. The line passing through the centers of the circles $\\omega_{1}$ and $\\omega_{2}$ is perpendicular to the line connecting the centers of the circles $\\omega_{1}$ and $\\omega_{2}$.\n6. The line connecting the centers of the circles $\\omega_{1}$ and $\\omega_{2}$ is perpendicular to the line passing through the centers of the circles $\\omega_{1}$ and $\\omega_{2}$.\n7. The line passing through the centers of the circles $\\omega_{1}$ and $\\omega_{2}$ is perpendicular to the line passing through the centers of the circles $\\omega_{1}$ and $\\omega_{2}$.\n8. The line passing through the centers of the circles $\\omega_{1}$ and $\\omega_{2}$ is perpendicular to the line passing through the centers of the circles $\\omega_{1}$ and $\\omega_{2}$.\n9. The line passing through the centers of the circles $\\omega_{1}$ and $\\omega_{2}$ is perpendicular to the line passing through the centers of the circles $\\omega_{1}$ and $\\omega_{2}$.\n10. The line passing through the centers of the circles $\\omega_{1}$ and $\\omega_{2}$ is perpendicular to the line passing through the centers of the circles $\\omega_{1}$ and $\\omega_{2}$.\n11. The line passing through the centers of the circles $\\omega_{1}$ and $\\omega_{2}$ is perpendicular to the line passing through the centers of the circles $\\omega_{1}$ and $\\omega_{2}$.\n12. The line passing through the centers of the circles $\\omega_{1}$ and $\\omega_{2}$ is perpendicular to the line passing through the centers of the circles $\\omega_{1}$ and $\\omega_{2}$.\n13. The line passing through the centers"}, {"kind": "sampled", "answer": "The relevant reasoning is: When points $K$, $M$, and $N$ are on the same line, it follows from the properties of circles that $\\angle KMN = 180^\\circ - \\angle MNK = 180^\\circ - \\angle KLM$ and $\\angle NML = 180^\\circ - \\angle LMN = 180^\\circ - \\angle LMK$.\n\nThis implies that line $MN$ bisects $\\angle LKM$, so $L$ and $M$ are on the same side of $MN$.\n\nSimilarly, line $MN$ bisects $\\angle LMN$, so $L$ and $N$ are on the same side of $MN$.\n\nTherefore, $K$, $M$, and $N$ are all on the same side of line $MN$.\n\nHence, the radius of the circle $\\omega$ is 3.\n\nFinal answer:\n#### 3"}]} +{"index": 64, "row": {"problem": "In triangle $ABC$, points $D$ and $E$ lie on side $AB$, with $D$ being closer to $A$. $AD=100 \\mathrm{~m}, EB=200 \\mathrm{~m}$, $\\angle ACD=30^{\\circ}, \\angle DCE=50^{\\circ}$, and $\\angle ECB=35^{\\circ}$. What is the length of segment $DE$?", "solution": "Let's introduce the following notations: $BC=a, CA=b, ED=x, EC=d, CD=e$, and let the height from point $C$ be $m$.\n\n![](https://cdn.mathpix.com/cropped/2024_05_02_585d0a62d800e11101d8g-1.jpg?height=689&width=1347&top_left_y=260&top_left_x=389)\n\nWrite the double area of the three sub-triangles and the entire $\\triangle ABC$ in two ways: using two sides and the sine of the included angle, and using the base and the corresponding height, then set the expressions equal to each other:\n\n$$\n\\begin{aligned}\n200 m & =a d \\sin 35^{\\circ} \\\\\nx m & =d e \\sin 50^{\\circ} \\\\\n100 m & =b e \\sin 30^{\\circ} \\\\\n(300+x) m & =a b \\sin 115^{\\circ}\n\\end{aligned}\n$$\n\nForm the product of equations (2) and (4), and the product of equations (1) and (3) - i.e., the products of the sides opposite each other - then divide the former product by the latter (which we can clearly do, since no area is 0). Thus, the auxiliary variables $a, b, d, e$ drop out, and we obtain the following equation:\n\n$$\n\\begin{gathered}\n\\frac{x(300+x)}{20000}=\\frac{\\sin 50^{\\circ} \\cdot \\sin 115^{\\circ}}{\\sin 35^{\\circ} \\cdot \\sin 30^{\\circ}} \\\\\nx^{2}+300 x-40000 \\frac{\\sin 50^{\\circ} \\cdot \\sin 65^{\\circ}}{\\sin 35^{\\circ}}=0\n\\end{gathered}\n$$\n\nFrom the signs of the coefficients, we see that the equation has one positive and one negative root, and only the positive root fits our problem and thought process:\n\n$$\nx=-150+\\sqrt{22500+40000 \\frac{\\sin 50^{\\circ} \\sin 65^{\\circ}}{\\sin 35^{\\circ}}}=116.3 \\text{ m}.\n$$", "answer": "116.3", "source": "olympiads", "problem_type": "Geometry", "uuid": "ffac367c-963d-587c-8221-460747944eba", "difficulty_band": "hard", "correctness_count": 1}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\nIn triangle $ABC$, points $D$ and $E$ lie on side $AB$, with $D$ being closer to $A$. $AD=100 \\mathrm{~m}, EB=200 \\mathrm{~m}$, $\\angle ACD=30^{\\circ}, \\angle DCE=50^{\\circ}$, and $\\angle ECB=35^{\\circ}$. What is the length of segment $DE$?", "candidates": [{"kind": "greedy", "answer": "To solve this problem, we can use the Law of Sines.\n\nFirst, we can find the length of side $AC$ using the Law of Sines:\n\n$$AC = 2R \\sin \\angle BAC$$\n\nwhere $R$ is the circumradius of the triangle.\n\nNow, we can use the Law of Sines to find the length of side $BC$:\n\n$$BC = 2R \\sin \\angle BAC$$\n\nNext, we can use the Law of Sines to find the length of side $AB$:\n\n$$AB = 2R \\sin \\angle ABC$$\n\nNow, we can use the Law of Sines to find the length of side $BC$:\n\n$$BC = 2R \\sin \\angle ABC$$\n\nNow, we can use the Law of Sines to find the length of side $AC$:\n\n$$AC = 2R \\sin \\angle BAC$$\n\nNow, we can use the Law of Sines to find the length of side $BC$:\n\n$$BC = 2R \\sin \\angle ABC$$\n\nNow, we can use the Law of Sines to find the length of side $AB$:\n\n$$AB = 2R \\sin \\angle ABC$$\n\nNow, we can use the Law of Sines to find the length of side $AC$:\n\n$$AC = 2R \\sin \\angle BAC$$\n\nNow, we can use the Law of Sines to find the length of side $BC$:\n\n$$BC = 2R \\sin \\angle ABC$$\n\nNow, we can use the Law of Sines to find the length of side $AB$:\n\n$$AB = 2R \\sin \\angle ABC$$\n\nNow, we can use the Law of Sines to find the length of side $AC$:\n\n$$AC = 2R \\sin \\angle BAC$$\n\nNow, we can use the Law of Sines to find the length of side $BC$:\n\n$$BC = 2R \\sin \\angle ABC$$\n\nNow, we can use the Law of Sines to find the length of side $AB$:\n\n$$AB = 2R \\sin \\angle ABC$$\n\nNow, we can use the Law of Sines"}, {"kind": "sampled", "answer": "The solution follows from the properties of the circle.\n\nFinal answer:\n#### 200"}]} +{"index": 65, "row": {"problem": "A college math class has $N$ teaching assistants. It takes the teaching assistants $5$ hours to grade homework assignments. One day, another teaching assistant joins them in grading and all homework assignments take only $4$ hours to grade. Assuming everyone did the same amount of work, compute the number of hours it would take for $1$ teaching assistant to grade all the homework assignments.", "solution": "1. Let \\( W \\) be the total amount of work required to grade all the homework assignments.\n2. If there are \\( N \\) teaching assistants, and it takes them 5 hours to grade all the homework, then the work done by each teaching assistant per hour is:\n \\[\n \\frac{W}{N \\cdot 5}\n \\]\n3. When another teaching assistant joins, making it \\( N+1 \\) teaching assistants, it takes them 4 hours to grade all the homework. Thus, the work done by each teaching assistant per hour is:\n \\[\n \\frac{W}{(N+1) \\cdot 4}\n \\]\n4. Since the total work \\( W \\) remains the same, we can set up the equation:\n \\[\n \\frac{W}{N \\cdot 5} = \\frac{W}{(N+1) \\cdot 4}\n \\]\n5. Simplifying the equation by canceling \\( W \\) from both sides, we get:\n \\[\n \\frac{1}{N \\cdot 5} = \\frac{1}{(N+1) \\cdot 4}\n \\]\n6. Cross-multiplying to solve for \\( N \\):\n \\[\n 4N = 5(N + 1)\n \\]\n7. Expanding and simplifying:\n \\[\n 4N = 5N + 5 \\implies 4N - 5N = 5 \\implies -N = 5 \\implies N = -5\n \\]\n8. Since \\( N \\) must be a positive integer, we re-evaluate the equation:\n \\[\n 4N = 5(N + 1) \\implies 4N = 5N + 5 \\implies N = 5\n \\]\n9. Therefore, there are \\( N = 4 \\) teaching assistants in the class.\n10. If 4 teaching assistants take 5 hours to grade the homework, the total work \\( W \\) can be calculated as:\n \\[\n W = 4 \\cdot 5 = 20 \\text{ hours}\n \\]\n11. Thus, it would take one teaching assistant 20 hours to grade all the homework assignments.\n\nThe final answer is \\( \\boxed{20} \\) hours.", "answer": "20", "source": "aops_forum", "problem_type": "Algebra", "uuid": "e9a9fa54-2709-5cb3-b471-7203d52cb4c6", "difficulty_band": "medium", "correctness_count": 2}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\nA college math class has $N$ teaching assistants. It takes the teaching assistants $5$ hours to grade homework assignments. One day, another teaching assistant joins them in grading and all homework assignments take only $4$ hours to grade. Assuming everyone did the same amount of work, compute the number of hours it would take for $1$ teaching assistant to grade all the homework assignments.", "candidates": [{"kind": "greedy", "answer": "The number of hours it would take for one teaching assistant to grade all the homework assignments is $4 \\times 1 = 4$ hours.\n\nFinal answer:\n#### 4"}, {"kind": "sampled", "answer": "The number of hours it would take for a single teaching assistant to grade all the homework assignments is 4 hours.\n\nFinal answer:\n#### 4"}]} +{"index": 66, "row": {"problem": "1. Find the smallest positive integer $k$, such that there exist positive integers $m, n$, satisfying $k=19^{n}-5^{m}$.", "solution": "1. Solution: $k_{\\text {min }}=14$ Hint: On one hand, take $m=n=1$, then $k=14$. On the other hand, let $k=19^{n}-5^{m}$.\n\nModulo 2 gives: $k \\equiv 0(\\bmod 2)$;\nModulo 3 gives: $k \\equiv 1^{n}-(-1)^{m} \\equiv 0$ or $2(\\bmod 3)$;\nModulo 5 gives: $k \\equiv(-1)^{n}(\\bmod 5)$.\nBased on the above conditions, we can determine that $k \\neq 1,2,3,4,5,7,8,9,10,11,12,13$.\nIf $k=6$, then modulo 19 gives $5^{m} \\equiv 13(\\bmod 19)$, which is impossible (by checking $5^{1}, 5^{2}, \\cdots, 5^{9}$, we find they are congruent to $5,6,11,17,9,7,16,4,1(\\bmod 19))$, so $k \\geqslant 14$.\nCombining both aspects, the minimum value of $k$ is 14.", "answer": "14", "source": "olympiads", "problem_type": "Number Theory", "uuid": "be9c969d-6432-505c-8bec-5b10a43ae2fd", "difficulty_band": "medium", "correctness_count": 2}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\n1. Find the smallest positive integer $k$, such that there exist positive integers $m, n$, satisfying $k=19^{n}-5^{m}$.", "candidates": [{"kind": "greedy", "answer": "To find the smallest positive integer $k$ such that there exist positive integers $m, n$, satisfying $k=19^{n}-5^{m}$, we need to find the smallest $k$ that can be expressed in the form $19^{n}-5^{m}$.\n\nSince $19^{n}-5^{m}$ is an integer, we can write it as $19^{n}-5^{m}=19^{n}-5^{n}=19^{n}-5^{n}=19^{n}-5^{n}=19^{n}-5^{n}=19^{n}-5^{n}=19^{n}-5^{n}=19^{n}-5^{n}=19^{n}-5^{n}=19^{n}-5^{n}=19^{n}-5^{n}=19^{n}-5^{n}=19^{n}-5^{n}=19^{n}-5^{n}=19^{n}-5^{n}=19^{n}-5^{n}=19^{n}-5^{n}=19^{n}-5^{n}=19^{n}-5^{n}=19^{n}-5^{n}=19^{n}-5^{n}=19^{n}-5^{n}=19^{n}-5^{n}=19^{n}-5^{n}=19^{n}-5^{n}=19^{n}-5^{n}=19^{n}-5^{n}=19^{n}-5^{n}=19^{n}-5^{n}=19^{n}-5^{n}=19^{n}-5^{n}=19^{n}-5^{n}=19^{n}-5^{n}=19^{n}-5^{n}=19^{n}-5^{n}=19^{n}-5^{n}=19^{n}-5^{n}=19^{n}-5^{n}=19^{n}-5^{n}=19^{n}-5^{n}=19^{n}-5^{n}=19^{n}-5^{n}=19^{n}-5^{n"}, {"kind": "sampled", "answer": "To find the smallest positive integer $k$ such that there exist positive integers $m, n$, satisfying $k=19^{n}-5^{m}$, we start by considering the properties of 19 and 5.\n\n19 is a power of 19, specifically $19 = 19^1$. This means we can express 19 as a power of 19 in the form $19 = m^a$, where $m$ and $a$ are positive integers.\n\n19 can be written as $19 = m^1$ where $m$ is a power of 19.\n\nWe want to find the smallest positive integer $k$ such that $k = 19^n - 5^m$ can be expressed as a power of 19.\n\nNotice that $19^n = (19^1)^n = 19^n = 19^1$. This means we can express 19^n as a power of 19 in the form $19^n = 19^1$.\n\nSo, the equation becomes $19^n = 19$. This equation has no positive integer solutions for $n$ and $a$, since $19^n$ is always greater than 1.\n\nHowever, we can see that $19 = 19^0$ is a valid solution.\n\nTherefore, the smallest positive integer $k$ is $k = \\boxed{1}$."}]} +{"index": 67, "row": {"problem": "2008 persons take part in a programming contest. In one round, the 2008 programmers are divided into two groups. Find the minimum number of groups such that every two programmers ever be in the same group.", "solution": "**\n - We need to show that 11 rounds are sufficient.\n - Number the 2008 programmers using 11-digit binary numbers from 0 to 2047 (adding leading zeros if necessary).\n - In each round \\( i \\), divide the programmers into teams based on the \\( i \\)-th digit of their binary number.\n - This ensures that in each round, the teams are balanced and every pair of programmers will be in different teams in at least one round because their binary numbers differ in at least one digit.\n\n4. **Verification:**\n - Consider any two programmers. Their binary representations differ in at least one digit.\n - Therefore, in the round corresponding to that digit, they will be in different teams.\n - This ensures that after 11 rounds, every pair of programmers has been in different teams at least once.\n\n5. **Inductive Proof:**\n - Base Case: For \\( k = 1 \\), with 2 programmers, 1 round is sufficient.\n - Inductive Step: Assume the claim is true for \\( k \\). For \\( 2^k < 2n \\leq 2^{k+1} \\), divide the programmers into two groups of sizes \\( 2^k \\) and \\( 2n - 2^k \\).\n - By the induction hypothesis, each group can be divided into equal-sized teams in \\( k \\) rounds.\n - For the \\( k+1 \\)-th round, swap teams to ensure that programmers from different groups have been in different teams at least once.\n - This completes the induction, proving that \\( k+1 \\) rounds suffice for \\( 2^k < 2n \\leq 2^{k+1} \\).\n\n\\(\\blacksquare\\)\n\nThe final answer is \\( \\boxed{ 11 } \\)", "answer": "11", "source": "aops_forum", "problem_type": "Combinatorics", "uuid": "b7eb61f0-a6f6-5884-8de6-21e9efc12b54", "difficulty_band": "medium", "correctness_count": 2}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\n2008 persons take part in a programming contest. In one round, the 2008 programmers are divided into two groups. Find the minimum number of groups such that every two programmers ever be in the same group.", "candidates": [{"kind": "greedy", "answer": "To find the minimum number of groups, we need to divide the total number of people by the number of people in each group.\n\nLet's denote the number of groups as $n$. We want to find the smallest $n$ such that $n$ is at least 2008.\n\nThe total number of people is $2008n$.\n\nThe number of people in each group is $n$.\n\nWe want to find the smallest $n$ such that $n \\geq 2008$.\n\nWe can write this as:\n\\[ n \\geq 2008 \\]\n\nThe smallest integer greater than or equal to 2008 is 2009.\n\nTherefore, the minimum number of groups is:\n\\[ n = \\boxed{2009} \\]"}, {"kind": "sampled", "answer": "2008(n-1) must divide 100400.\n\nThe prime factorization of 2008 is 2^4 * 3^2 * 167.\n\nThe minimum number of groups that is enough to satisfy the condition is 2 * 5 = 10.\n\nFinal answer:\n#### 10"}]} +{"index": 68, "row": {"problem": "7.242. $\\left(16 \\cdot 5^{2 x-1}-2 \\cdot 5^{x-1}-0.048\\right) \\lg \\left(x^{3}+2 x+1\\right)=0$.", "solution": "Solution.\n\nDomain: $x^{3}+2 x+1>0$.\n\nFrom the condition $16 \\cdot 5^{2 x-1}-2^{x-1}-0.048=0$ or $\\lg \\left(x^{3}+2 x+1\\right)=0$. Rewrite the first equation as\n\n$\\frac{16}{5} \\cdot 5^{2 x}-\\frac{2}{5} \\cdot 5^{x}-0.048=0 \\Leftrightarrow 16 \\cdot 5^{2 x}-2 \\cdot 5^{x}-0.24=0$.\n\nSolving this equation as a quadratic in terms of $5^{x}$, we get $5^{x}=-\\frac{3}{40}$ (no solutions), or $5^{x}=5^{-1} \\Leftrightarrow x_{1}=-1$ (does not satisfy the domain). From the second equation, we have\n\n$x^{3}+2 x+1=1 \\Leftrightarrow x^{3}+2 x=0 \\Leftrightarrow x\\left(x^{2}+2\\right)=0, x_{3}=0, x^{2}+2 \\neq 0$.\n\nAnswer: 0.", "answer": "0", "source": "olympiads", "problem_type": "Algebra", "uuid": "2bfb42a5-11c9-5aee-b6df-edb9b8efee0e", "difficulty_band": "medium", "correctness_count": 2}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\n7.242. $\\left(16 \\cdot 5^{2 x-1}-2 \\cdot 5^{x-1}-0.048\\right) \\lg \\left(x^{3}+2 x+1\\right)=0$.", "candidates": [{"kind": "greedy", "answer": "1. Start by rewriting the given equation:\n\n$$\\left(16 \\cdot 5^{2x-1}-2 \\cdot 5^{x-1}-0.048\\right) \\lg \\left(x^{3}+2 x+1\\right)=0$$\n\n2. Use the property of logarithms that $\\lg a^b = b \\lg a$ to simplify the left-hand side:\n\n$$\\left(16 \\cdot 5^{2x-1}-2 \\cdot 5^{x-1}-0.048\\right) \\lg \\left(x^{3}+2 x+1\\right)=0$$\n\n3. Rewrite the logarithmic expression using the property $\\lg a^b = b \\lg a$:\n\n$$\\left(16 \\cdot 5^{2x-1}-2 \\cdot 5^{x-1}-0.048\\right) \\lg \\left(x^{3}+2 x+1\\right)=0$$\n\n4. Since the logarithm is one-to-one, we can equate the arguments of the logarithms:\n\n$$16 \\cdot 5^{2x-1}-2 \\cdot 5^{x-1}-0.048 = 0$$\n\n5. Rearrange the equation to isolate the exponential term:\n\n$$16 \\cdot 5^{2x-1} = 2 \\cdot 5^{x-1}$$\n\n6. Divide both sides by $2 \\cdot 5^{x-1}$:\n\n$$\\left(16 \\cdot 5^{2x-1}\\right) / (2 \\cdot 5^{x-1}) = 1$$\n\n7. Simplify the left-hand side:\n\n$$\\left(16 \\cdot 5^{2x-1}\\right) / (2 \\cdot 5^{x-1}) = 1$$\n\n8. Cancel out the common factor of $2 \\cdot 5^{x-1}$:\n\n$$\\left(16 \\cdot 5^{2x-1}\\right) / (2 \\cdot 5^{x-1}) = 1$$\n\n9. Simplify the left-hand side:\n\n$$\\left(1"}, {"kind": "sampled", "answer": "11629122286. The final answer is 0"}]} +{"index": 69, "row": {"problem": "Example 1. Find the residues at the singular points of the function\n\n$$\nf(z)=\\frac{\\tan z}{z^{2}-\\pi z / 4}\n$$", "solution": "SOLUTION. Let's represent the function $f(z)$ as\n\n$$\nf(z)=\\frac{\\operatorname{tg} z}{z^{2}-\\pi z / 4}=\\frac{\\sin z}{z \\cos z(z-\\pi / 4)}\n$$\n\n1. We find the isolated singular points of the function $f(z)$. The singular points of the function are the zeros of the denominator, i.e.\n\n$$\nz=0, \\quad z=\\pi / 4, \\quad z_{k}=\\frac{\\pi}{2}+\\pi k, \\quad k=0, \\pm 1, \\ldots\n$$\n\n2. We determine the type of each singular point and compute the residue at each:\n\n$z=0$ - a removable singular point, since\n\n$$\n\\lim _{z \\rightarrow 0} f(z)=\\lim _{z \\rightarrow 0} \\frac{\\sin z}{z \\cos z(z-\\pi / 4)}=-\\frac{4}{\\pi}\n$$\n\nTherefore, by formula (1)\n\n$$\n\\operatorname{res}_{z=0} f(z)=0 \\text {. }\n$$\n\nThe point $z=\\pi / 4$ is a simple pole, so by formula (3)\n\n$$\n\\begin{aligned}\n& \\operatorname{res}_{z=\\pi / 4} f(z)=\\lim _{z \\rightarrow \\pi / 4}(z-\\pi / 4) f(z)= \\\\\n& =\\lim _{z \\rightarrow \\pi / 4} \\frac{(z-\\pi / 4) \\operatorname{tg} z}{z(z-\\pi / 4)}=\\lim _{z \\rightarrow \\pi / 4} \\frac{\\operatorname{tg} z}{z}=4 / \\pi\n\\end{aligned}\n$$\n\nThe points $z_{k}=\\pi / 2+\\pi k$ are simple poles of the function $f(z)$. Let's represent $f(z)$ as\n\n$$\nf(z)=\\frac{\\frac{\\sin z}{z^{2}-\\pi z / 4}}{\\cos z}=\\frac{\\varphi(z)}{\\psi(z)}\n$$\n\nwhere $\\varphi(z)$ and $\\psi(z)$ are analytic at the points $z_{k}=\\pi / 2+\\pi k$ and\n\n$$\n\\varphi\\left(z_{k}\\right)=\\frac{\\sin z_{k}}{z_{k}^{2}-\\pi z_{k} / 4} \\neq 0, \\quad \\psi\\left(z_{k}\\right)=\\cos z_{k}=0, \\quad \\psi^{\\prime}\\left(z_{k}\\right)=-\\sin z_{k} \\neq 0\n$$\n\nTherefore, by formula (4)\n\n$$\n\\begin{aligned}\n& \\operatorname{res}_{z=z_{k}} f(z)=\\frac{\\frac{\\sin z_{k}}{z_{k}\\left(z_{k}-\\pi / 4\\right)}}{-\\sin z_{k}}= \\\\\n& =\\frac{\\frac{\\sin (\\pi / 2+\\pi k)}{(\\pi / 2+\\pi k)(\\pi / 2+\\pi k-\\pi / 4)}}{-\\sin (\\pi / 2+\\pi k)}=-\\frac{1}{(\\pi / 2+\\pi k)(\\pi / 4+\\pi k)} . \\\\\n& \\text { Answer. } \\operatorname{res}_{z=0} \\frac{\\operatorname{tg} z}{z^{2}-\\pi z / 4}=0, \\quad \\operatorname{res}_{z=\\pi / 4} \\frac{\\operatorname{tg} z}{z^{2}-\\pi z / 4}=\\frac{4}{\\pi} \\\\\n& \\operatorname{res}_{z=\\pi / 2+\\pi k} \\frac{\\operatorname{tg} z}{z^{2}-\\pi z / 4}=-\\frac{1}{(\\pi / 2+\\pi k)(\\pi / 4+\\pi k)} .\n\\end{aligned}\n$$", "answer": "\\operatorname{res}_{0}\\frac{\\operatorname{tg}z}{z^{2}-\\piz/4}=0,\\quad\\operatorname{res}_{\\pi/4}\\frac{\\operatorname{tg}z}{z^{2}-\\piz/4}=\\frac{4}{\\pi},\\quad\\operatorname{res}_{\\pi/2+\\pik}\\frac", "source": "olympiads", "problem_type": "Calculus", "uuid": "a14f50b6-4fe6-5116-9c74-a0361349d083", "difficulty_band": "hard", "correctness_count": 1}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\nExample 1. Find the residues at the singular points of the function\n\n$$\nf(z)=\\frac{\\tan z}{z^{2}-\\pi z / 4}\n$$", "candidates": [{"kind": "greedy", "answer": "The function is defined as:\n$$\nf(z)=\\frac{\\tan z}{z^{2}-\\pi z / 4}\n$$\n\nWe need to find the residues at the singular points of the function.\n\nFirst, we need to find the zeros of the denominator $z^{2}-\\pi z / 4$.\n\nFactor the denominator:\n$$\nz^{2}-\\pi z / 4 = z(z-1/4)\n$$\n\nThe zeros of the denominator are at $z = 0$ and $z = 1/4$.\n\nNext, we need to find the zeros of the numerator $z^{2}-\\pi z / 4$.\n\nFactor the numerator:\n$$\nz^{2}-\\pi z / 4 = z(z-1/4)\n$$\n\nThe zeros of the numerator are at $z = 0$ and $z = 1/4$.\n\nNow, we need to find the residues at the singular points of the function.\n\nThe singular points are at $z = 0$ and $z = 1/4$.\n\nThe residues at these points are:\n$$\nf(0) = \\frac{\\tan 0}{0^{2}-\\pi \\cdot 0 / 4} = 0\n$$\n$$\nf(1/4) = \\frac{\\tan (1/4)}{(1/4)^{2}-\\pi \\cdot (1/4) / 4} = 1\n$$\n\nThe residues at the singular points are 0 and 1.\n\nFinal answer:\n#### 0"}, {"kind": "sampled", "answer": "To find the residues at the singular points of the function $f(z)=\\frac{\\tan z}{z^{2}-\\pi z/4}$, we need to find the values of $z$ for which the function is undefined. The function is undefined when the denominator $z^{2}-\\pi z/4$ equals 0.\n\nSolve the equation $z^{2}-\\pi z/4=0$ for $z$. This is done by factoring the quadratic equation:\n\n$$z^{2}-\\pi z/4 = 0$$\n$$z^{2} - \\pi z / 4 = 0$$\n\nFactor out $z$:\n\n$$z(z - \\pi / 4) = 0$$\n\nSolve for $z$:\n\n$$z = 0, \\pi / 4$$\n\nSo, the function $f(z)$ has three singular points at $z = 0, \\pi / 4$.\n\nThe answer is:\n$\\boxed{0, \\pi/4}$"}]} +{"index": 70, "row": {"problem": "Question 15: Let the set $M=\\{1,2, \\ldots, 100\\}$ be a 100-element set. If for any n-element subset $A$ of $M$, there are always 4 elements in $A$ that are pairwise coprime, find the minimum value of $\\mathrm{n}$.", "solution": "Question 15, Solution: Let $A=\\{x|x \\in M, 2| x\\} \\cup\\{x|x \\in M, 3| x\\} \\cup\\{x|x \\in M, 5| x\\}$, then according to the principle of inclusion-exclusion, we have:\n$$\n|A|=\\left[\\frac{100}{2}\\right]+\\left[\\frac{100}{3}\\right]+\\left[\\frac{100}{5}\\right]-\\left[\\frac{100}{2 \\times 3}\\right]-\\left[\\frac{100}{2 \\times 5}\\right]-\\left[\\frac{100}{3 \\times 5}\\right]+\\left[\\frac{100}{2 \\times 3 \\times 5}\\right]=74\n$$\n\nSince every element in $A$ is a multiple of 2, 3, or 5, any four elements chosen from $A$ must include at least two elements that are not coprime.\n$$\n\\begin{array}{l}\n\\text { On the other hand, when } n=75, \\text { we can take } \\\\\nA_{1}=\\{1\\} \\cup\\{\\text { all primes not exceeding } 100\\} ; \\\\\nA_{2}=\\left\\{2^{2}, 3^{2}, 5^{2}, 7^{2}\\right\\} ; \\\\\nA_{3}=\\left\\{2^{3}, 3^{3}, 5 \\times 19,7 \\times 13\\right\\} ; \\\\\nA_{4}=\\left\\{2^{4}, 3^{4}, 5 \\times 17,7 \\times 11\\right\\}\n\\end{array}\n$$\n\nThen all the numbers in these four sets are distinct, and any two numbers in each set are coprime. These four sets contain a total of 38 numbers. When 75 numbers are chosen from $M$, at least $75-(100-38)=13$ numbers must come from these four subsets. By the pigeonhole principle, at least one of these subsets must contain $\\left\\lceil\\frac{13}{4}\\right\\rceil=4$ numbers, meaning these 4 numbers are pairwise coprime. In summary, the smallest value of $n$ is 75.", "answer": "75", "source": "olympiads", "problem_type": "Combinatorics", "uuid": "15e04de3-b903-5f2c-b8fa-0e8a23a495f3", "difficulty_band": "medium", "correctness_count": 2}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\nQuestion 15: Let the set $M=\\{1,2, \\ldots, 100\\}$ be a 100-element set. If for any n-element subset $A$ of $M$, there are always 4 elements in $A$ that are pairwise coprime, find the minimum value of $\\mathrm{n}$.", "candidates": [{"kind": "greedy", "answer": "To find the minimum value of $n$ such that for any $n$-element subset $A$ of $M$, there are always 4 elements in $A$ that are pairwise coprime, we need to consider the prime factorization of $M$.\n\nThe prime factorization of $M$ is $2^3 \\cdot 3^2 \\cdot 5^1$.\n\nTo ensure that any $n$-element subset $A$ of $M$ has 4 elements that are pairwise coprime, we need to consider the following:\n\n* For any $n$, we can choose $n$ elements from $M$ such that they are pairwise coprime.\n* We can choose $n$ elements from $M$ such that they are not pairwise coprime.\n\nThe smallest $n$ that satisfies this condition is $n = 100$.\n\nTherefore, the minimum value of $n$ is $\\boxed{100}$.\nFinal answer:\n#### 100"}, {"kind": "sampled", "answer": "The smallest number that is not coprime with 100 is 123, which is the number of prime numbers less than or equal to 100.\n\nTherefore, the minimum value of $n$ is 123.\n\nThe final answer is:\n#### 123"}]} +{"index": 71, "row": {"problem": "A sand pit is constructed in the shape of a rectangular prism $10 \\mathrm{~m}$ long, $50 \\mathrm{~cm}$ deep and $2 \\mathrm{~m}$ wide. If the pit is already half-full, how much more sand, in $\\mathrm{m}^{3}$, is needed to completely fill the pit?\n(A) 6\n(B) 5\n(C) 20\n(D) 7.5\n(E) 10", "solution": "The volume of the entire pit is\n\n$$\n(10 \\mathrm{~m}) \\times(50 \\mathrm{~cm}) \\times(2 \\mathrm{~m})=(10 \\mathrm{~m}) \\times(0.5 \\mathrm{~m}) \\times(2 \\mathrm{~m})=10 \\mathrm{~m}^{3}\n$$\n\nSince the pit starts with $5 \\mathrm{~m}^{3}$ in it, an additional $5 \\mathrm{~m}^{3}$ of sand is required to fill it.\n\nANswer: (B)", "answer": "5", "source": "olympiads", "problem_type": "Geometry", "uuid": "9067b264-f906-5845-8f1a-c993d3c58037", "difficulty_band": "hard", "correctness_count": 1}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\nA sand pit is constructed in the shape of a rectangular prism $10 \\mathrm{~m}$ long, $50 \\mathrm{~cm}$ deep and $2 \\mathrm{~m}$ wide. If the pit is already half-full, how much more sand, in $\\mathrm{m}^{3}$, is needed to completely fill the pit?\n(A) 6\n(B) 5\n(C) 20\n(D) 7.5\n(E) 10", "candidates": [{"kind": "greedy", "answer": "The volume of the sand pit is given by the formula:\n\n$$V = lwh$$\n\nwhere $l$ is the length, $w$ is the width, and $h$ is the height of the pit.\n\nSubstituting the given values for the length, width, and height, we get:\n\n$$V = 10 \\, m \\times 50 \\, cm \\times 2 \\, m$$\n\nNow, plug in the given values for the length, width, and height to calculate the volume of the pit:\n\n$$V = 10 \\times 50 \\times 2$$\n\nCalculate the volume of the pit:\n\n$$V = 10 \\times 50 \\times 2 = 1000 \\, m^3$$\n\nNow, we need to find out how much more sand is needed to fill the pit. We can do this by subtracting the volume of the pit from the volume of the pit that is already full:\n\n$$Additional\\, sand = V - V$$\n\nPlug in the values for the volume of the pit and the volume of the pit that is already full to calculate the additional sand:\n\n$$Additional\\, sand = 1000 - 1000$$\n\nCalculate the additional sand needed:\n\n$$Additional\\, sand = 1000 - 1000 = 0 \\, m^3$$\n\nSo, 0 m^3 of sand is needed to completely fill the pit.\n\nThe answer is:\n\\boxed{0}"}, {"kind": "sampled", "answer": "Given:\n- Length: $10 \\, m$\n- Width: $50 \\, cm$\n- Depth: $2 \\, m$\n- Water level: $\\frac{10}{2} = 5 \\, m$\n\nWe can find the volume of water required to fill the pit by multiplying the volume of the pit by the depth of water.\n\n$$Volume = \\text{Volume} \\\\times \\text{Depth}$$\n\nSubstituting the given values:\n\n$$Volume = (10 \\, m \\times 5 \\, cm) \\\\times 2 \\, m$$\n\nCalculate the volume of water needed:\n\n$$Volume = (10 \\times 5) \\, cm \\\\times 2 \\, m = 50 \\, cm \\times 2 \\, m = 100 \\, m^3$$\n\nTherefore, you need $100 \\, m^3$ of sand to completely fill the pit.\n\nFinal answer:\n#### 100"}]} +{"index": 72, "row": {"problem": "How many integers between $123$ and $789$ have at least two identical digits, when written in base $10?$\n$$\n\\mathrm a. ~ 180\\qquad \\mathrm b.~184\\qquad \\mathrm c. ~186 \\qquad \\mathrm d. ~189 \\qquad \\mathrm e. ~191\n$$", "solution": "To determine how many integers between $123$ and $789$ have at least two identical digits, we can follow these steps:\n\n1. **Calculate the total number of integers between $123$ and $789$:**\n \\[\n 789 - 123 + 1 = 667\n \\]\n\n2. **Calculate the number of integers between $123$ and $789$ with all distinct digits:**\n - We need to count the number of three-digit numbers with distinct digits in the range $123$ to $789$.\n - A three-digit number $\\overline{abc}$ has distinct digits if $a, b, c$ are all different.\n\n3. **Count the number of three-digit numbers with distinct digits:**\n - The first digit $a$ can be any digit from $1$ to $7$ (since $789$ is the upper limit).\n - The second digit $b$ can be any digit from $0$ to $9$ except $a$.\n - The third digit $c$ can be any digit from $0$ to $9$ except $a$ and $b$.\n\n Therefore, the number of distinct three-digit numbers $\\overline{abc}$ is:\n \\[\n 7 \\times 9 \\times 8 = 504\n \\]\n\n4. **Calculate the number of integers with at least two identical digits:**\n - Subtract the number of integers with all distinct digits from the total number of integers.\n \\[\n 667 - 504 = 163\n \\]\n\n5. **Adjust for the range $123$ to $789$:**\n - The solution provided in the community has a different approach, which involves calculating the number of integers with at least two identical digits directly.\n - According to the solution, the number of integers in $[100, 789]$ with at least two repeated digits is:\n \\[\n 28 \\times 7 + 2 \\times 8 - 18 = 194\n \\]\n - The number of integers in $[100, 122]$ with at least two repeated digits is:\n \\[\n 28 \\times 1 + 2 \\times 2 - 18 = 14\n \\]\n - Therefore, the number of integers in $[123, 789]$ with at least two repeated digits is:\n \\[\n 194 - 14 = 180\n \\]\n\nThe final answer is $\\boxed{180}$", "answer": "180", "source": "aops_forum", "problem_type": "Combinatorics", "uuid": "7525fb6b-2ced-5c91-92f6-d48e9c5b381f", "difficulty_band": "hard", "correctness_count": 1}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\nHow many integers between $123$ and $789$ have at least two identical digits, when written in base $10?$\n$$\n\\mathrm a. ~ 180\\qquad \\mathrm b.~184\\qquad \\mathrm c. ~186 \\qquad \\mathrm d. ~189 \\qquad \\mathrm e. ~191\n$$", "candidates": [{"kind": "greedy", "answer": "To find the number of integers between 123 and 789 (inclusive) that have at least two identical digits when written in base 10, we can use the principle of inclusion-exclusion.\n\nFirst, we need to find the total number of integers between 123 and 789, inclusive. This is the difference between the two numbers, which is 789 - 123 = 666.\n\nNext, we need to find the number of integers that have at least two identical digits. We can do this by subtracting the number of integers that have no identical digits from the total number of integers.\n\nLet's denote the number of integers with no identical digits as $N$. Then, the number of integers with at least two identical digits is $N - N + N - 1 = 3N - 2$.\n\nNow, we can plug in the given values for $N$ and $N - 1$ to calculate the number of integers with at least two identical digits:\n\n$$3(123) - 2 = 369$$\n\nSo, there are 369 integers between 123 and 789 (inclusive) that have at least two identical digits when written in base 10.\n\nThe answer is:\n\\[ \\boxed{369} \\]\n\nFinal answer:\n#### 369"}, {"kind": "sampled", "answer": "First, we need to find the smallest and largest integers between $123$ and $789$ in base $10$. We'll do this by converting the given values into base $10$.\n\n$$123_{10} = 1 \\times 10^2 + 2 \\times 10^1 + 3 \\times 10^0 = 123$$\n\n$$789_{10} = 7 \\times 10^2 + 8 \\times 10^1 + 9 \\times 10^0 = 789$$\n\nNow, let's find the integers between $123$ and $789$ in base $10$. We'll do this by considering the digits in each place value (left to right).\n\n$$\\begin{array}{c|c}\n\\text{Digit} & \\text{Base} \\\\ \\hline\n1 & 2 \\\\ 123_{10} & 12 \\\\ 789_{10} & 789 \\\\ \\hline\n2 & 1 \\\\ 123_{10} & 11 \\\\ 789_{10} & 789 \\\\ \\hline\n3 & 0 \\\\ 123_{10} & 10 \\\\ 789_{10} & 789 \\\\ \\hline\n4 & 0 \\\\ 123_{10} & 12 \\\\ 789_{10} & 789 \\\\ \\hline\n5 & 0 \\\\ 123_{10} & 10 \\\\ 789_{10} & 789 \\\\ \\hline\n6 & 0 \\\\ 123_{10} & 11 \\\\ 789_{10} & 789 \\\\ \\hline\n7 & 1 \\\\ 123_{10} & 10 \\\\ 789_{10} & 789 \\\\ \\hline\n8 & 0 \\\\ 123_{10} & 12 \\\\ 789_{10} & 789 \\\\ \\hline\n9 & 0 \\\\ 123_{10} &"}]} +{"index": 73, "row": {"problem": "[ The Law of Sines [ Area of a Triangle (using two sides and the included angle).]\n\nOn a line passing through the center $O$ of a circle with radius 12, points $A$ and $B$ are taken such that $O A=15, A B=5$, and $A$ lies between $O$ and $B$. Tangents are drawn from points $A$ and $B$ to the circle, with the points of tangency lying on the same side of the line $O B$. Find the area of triangle $A B C$, where $C$ is the point of intersection of these tangents.", "solution": "Apply the Law of Sines to triangle $ABC$.\n\n## Solution\n\nLet $M$ and $N$ be the points of tangency of the circle with the lines passing through points $A$ and $B$ respectively,\n\n$\\angle O A M=\\alpha, \\angle O B N=\\beta$. Then\n\n$$\n\\frac{O M}{O A}=\\sin \\alpha, \\frac{O N}{O B}=\\sin \\beta\n$$\n\nTherefore,\n\n$$\n\\begin{gathered}\n\\sin \\alpha=\\frac{4}{5}, \\sin \\beta=\\frac{3}{5}, \\cos \\alpha=\\frac{3}{5}, \\cos \\beta=\\frac{4}{5} \\\\\nB C=\\frac{A B \\sin \\alpha}{\\sin (\\alpha-\\beta)}=\\frac{A B \\sin \\alpha}{\\sin \\alpha \\cos \\beta-\\cos \\alpha \\sin \\beta}=\\frac{100}{7}\n\\end{gathered}\n$$\n\nThus,\n\n$$\nS_{\\triangle \\mathrm{ABC}}=\\frac{1}{2} A B \\cdot B C \\sin \\angle A B C=\\frac{1}{2} \\cdot 5 \\cdot \\frac{100}{7} \\cdot \\frac{3}{5}=\\frac{150}{7}\n$$\n\n![](https://cdn.mathpix.com/cropped/2024_05_06_96a7587f2ba0a7b3aee2g-46.jpg?height=378&width=738&top_left_y=820&top_left_x=660)\n\n## Answer\n\n$\\frac{150}{7}$.", "answer": "\\frac{150}{7}", "source": "olympiads", "problem_type": "Geometry", "uuid": "4a02e6ca-98b0-5ec2-b2fa-ad5dda27a0be", "difficulty_band": "medium", "correctness_count": 2}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\n[ The Law of Sines [ Area of a Triangle (using two sides and the included angle).]\n\nOn a line passing through the center $O$ of a circle with radius 12, points $A$ and $B$ are taken such that $O A=15, A B=5$, and $A$ lies between $O$ and $B$. Tangents are drawn from points $A$ and $B$ to the circle, with the points of tangency lying on the same side of the line $O B$. Find the area of triangle $A B C$, where $C$ is the point of intersection of these tangents.", "candidates": [{"kind": "greedy", "answer": "The area of triangle $ABC$ is equal to the area of the circle with radius 12 multiplied by the ratio of the distance from the center of the circle to the line segment $AB$ to the length of the segment $AB$.\n\nThe distance from the center of the circle to the line segment $AB$ is 15.\n\nThe distance from the center of the circle to the line segment $AB$ is 5.\n\nTherefore, the ratio of the distance from the center of the circle to the line segment $AB$ to the length of the segment $AB$ is 5/15 = 1/3.\n\nTherefore, the area of triangle $ABC$ is 1/3 of the area of the circle with radius 12 multiplied by the ratio of the distance from the center of the circle to the line segment $AB$ to the length of the segment $AB$.\n\nTherefore, the area of triangle $ABC$ is 1/3 * 12 * (1/3) = 4.\n\nThe answer is:\n\\[ \\boxed{4} \\]"}, {"kind": "sampled", "answer": "Given that $OA=15$, $OB=5$, and $A$ lies on the circle, we can find the length of the chord $AB$. The length of chord $AB$ can be found using the formula for the length of a chord in a circle: $AB = 2r \\sin\\left(\\frac{\\theta}{2}\\right)$ where $r$ is the radius of the circle and $\\theta$ is the angle subtended by the chord at the center of the circle.\n\n\\[ AB = 2 \\cdot 12 \\cdot \\sin\\left(\\frac{\\pi}{5}\\right) = 24\\sqrt{2} \\]\n\nNow, let's denote the distance from point $C$ to line $OA$ as $t$. Since $OC=12$, we can write:\n\n\\[ t = \\frac{OA}{\\sin\\left(\\frac{\\angle A}{2}\\right)} = \\frac{15}{2\\sin\\left(\\frac{\\pi}{5}\\right)} = \\frac{15}{2 \\cdot \\frac{\\sqrt{2}}{2}} = \\frac{15\\sqrt{2}}{4} \\]\n\nThe area of triangle $ABC$ is given by:\n\n\\[ \\text{Area} = \\frac{1}{2} \\cdot AB \\cdot t = \\frac{1}{2} \\cdot 24\\sqrt{2} \\cdot \\frac{15\\sqrt{2}}{4} = \\boxed{105} \\]"}]} +{"index": 74, "row": {"problem": "7.5. One hundred non-zero integers are written in a circle such that each number is greater than the product of the two numbers following it in a clockwise direction. What is the maximum number of positive numbers that can be among these 100 written numbers?", "solution": "Answer: 50.\n\nSolution: Note that two consecutive numbers cannot both be positive (i.e., natural numbers). Suppose the opposite. Then their product is positive, and the number before them (counterclockwise) is also a natural number. Since it is greater than the product of these two natural numbers, it is greater than each of them. Continuing this reasoning, we will conclude that all numbers are natural, and if we move counterclockwise, the numbers will increase. However, when the \"circle closes,\" we will arrive at a contradiction.\n\nTherefore, in any pair of adjacent numbers, there is at least one negative number. Dividing our 100 numbers into 50 pairs, we get that there are no fewer than 50 negative numbers, and thus no more than 50 positive numbers. If we alternate a positive number 2 and a negative number -2, we will get an example of the arrangement of 50 positive and 50 negative numbers that satisfy the problem's condition.\n\nComment: Correct answer without justification - 0 points.\n\nProving the impossibility of placing two positive numbers next to each other - 3 points.\n\nProving that there are no more than 50 positive numbers - 1 point.\n\nProviding an example with 50 positive numbers - 3 points.", "answer": "50", "source": "olympiads", "problem_type": "Number Theory", "uuid": "4c70b2dc-7439-578e-98e5-f8e567012637", "difficulty_band": "medium", "correctness_count": 2}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\n7.5. One hundred non-zero integers are written in a circle such that each number is greater than the product of the two numbers following it in a clockwise direction. What is the maximum number of positive numbers that can be among these 100 written numbers?", "candidates": [{"kind": "greedy", "answer": "To maximize the number of positive numbers among the 100 non-zero integers written in a circle, we should arrange the numbers in a way that each number is as large as possible relative to the product of the two numbers following it.\n\nLet's denote the numbers as $a_1, a_2, \\ldots, a_{100}$, where $a_1 < a_2 < \\ldots < a_{100}$.\n\nWe want to maximize the product $a_1a_2\\ldots a_{100}$.\n\nSince each number is greater than the product of the two numbers following it in a clockwise direction, we can assume that $a_i > a_{i+1} \\cdot a_{i+2}$ for all $i$.\n\nTo maximize the product, we want to make $a_i$ as large as possible relative to $a_{i+1}$ and $a_{i+2}$.\n\nLet's consider the case where $a_i$ is as large as possible relative to $a_{i+1}$ and $a_{i+2}$.\n\nWe can do this by setting $a_i = a_{i+1} \\cdot a_{i+2}$.\n\nThis way, we can maximize the product $a_1a_2\\ldots a_{100}$.\n\nIn this case, we have $a_1 = a_{100}$, $a_2 = a_{101}$, and $a_{102} = a_{103}$.\n\nTherefore, the maximum number of positive numbers among the 100 non-zero integers written in a circle is $100$.\n\nThe answer is:\n\\[ \\boxed{100} \\]"}, {"kind": "sampled", "answer": "1. **Understanding the Problem:**\n - We are given 100 non-zero integers arranged in a circle.\n - Each integer is greater than the product of the two integers immediately following it in a clockwise direction.\n\n2. **Analyzing the Assuming:**\n - Let’s denote the first non-zero integer by \\( k \\).\n - Since the integers are in a circle, the following inequalities must hold:\n \\[\n k > k \\cdot p_1 \\quad \\text{and} \\quad k > k \\cdot p_2\n \\]\n where \\( p_1 \\) and \\( p_2 \\) are the next integers following \\( k \\).\n\n3. **Applying the Condition:**\n - From the inequalities above, we get:\n \\[\n k > kp_1 \\quad \\text{and} \\quad k > kp_2\n \\]\n - Dividing both sides of the first inequality by \\( k \\), we obtain:\n \\[\n 1 > p_1\n \\]\n - Similarly, dividing both sides of the second inequality by \\( k \\), we obtain:\n \\[\n 1 > p_2\n \\]\n\n4. **Finding the Maximum:**\n - Since \\( p_1 \\) and \\( p_2 \\) are positive integers, the maximum product of two positive integers \\( p_1 \\) and \\( p_2 \\) is achieved when \\( p_1 = p_2 \\).\n - Therefore, the maximum product of two positive integers in the set is \\( p_1 p_2 \\).\n\n5. **Maximizing the Number of Positive Numbers:**\n - The maximum possible product of two positive integers in the set is \\( p_1 p_2 \\).\n - To maximize the number of positive integers, we need to find the largest value of \\( p_1 p_2 \\) that is less than or equal to 100.\n\n6. **Finding the Maximum Value:**\n - Let's try to find the largest value of \\( p_1 p_2 \\) that is less than or equal to 100.\n - We can do this by trial and error or by using the quadratic formula.\n\n7. **Calculation for the Maximum Value:**\n - We can compute the values of \\( p_1 \\) and"}]} +{"index": 75, "row": {"problem": "Let $B = (20, 14)$ and $C = (18, 0)$ be two points in the plane. For every line $\\ell$ passing through $B$, we color red the foot of the perpendicular from $C$ to $\\ell$. The set of red points enclose a bounded region of area $\\mathcal{A}$. Find $\\lfloor \\mathcal{A} \\rfloor$ (that is, find the greatest integer not exceeding $\\mathcal A$).\n\n[i]Proposed by Yang Liu[/i]", "solution": "1. **Determine the distance \\( BC \\):**\n - The coordinates of points \\( B \\) and \\( C \\) are \\( B = (20, 14) \\) and \\( C = (18, 0) \\).\n - The distance formula between two points \\((x_1, y_1)\\) and \\((x_2, y_2)\\) is given by:\n \\[\n d = \\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\n \\]\n - Applying this formula to points \\( B \\) and \\( C \\):\n \\[\n BC = \\sqrt{(18 - 20)^2 + (0 - 14)^2} = \\sqrt{(-2)^2 + (-14)^2} = \\sqrt{4 + 196} = \\sqrt{200} = 10\\sqrt{2}\n \\]\n\n2. **Identify the set of red points:**\n - For every line \\(\\ell\\) passing through \\( B \\), the foot of the perpendicular from \\( C \\) to \\(\\ell\\) will trace out a circle centered at the midpoint of \\( BC \\) with radius equal to half the distance \\( BC \\).\n - The midpoint \\( M \\) of \\( BC \\) is:\n \\[\n M = \\left( \\frac{20 + 18}{2}, \\frac{14 + 0}{2} \\right) = (19, 7)\n \\]\n - The radius of the circle is:\n \\[\n r = \\frac{BC}{2} = \\frac{10\\sqrt{2}}{2} = 5\\sqrt{2}\n \\]\n\n3. **Calculate the area of the circle:**\n - The area \\( \\mathcal{A} \\) of a circle with radius \\( r \\) is given by:\n \\[\n \\mathcal{A} = \\pi r^2\n \\]\n - Substituting \\( r = 5\\sqrt{2} \\):\n \\[\n \\mathcal{A} = \\pi (5\\sqrt{2})^2 = \\pi \\cdot 25 \\cdot 2 = 50\\pi\n \\]\n\n4. **Find the greatest integer not exceeding \\( \\mathcal{A} \\):**\n - Using the approximation \\( \\pi \\approx 3.14159 \\):\n \\[\n 50\\pi \\approx 50 \\times 3.14159 = 157.0795\n \\]\n - The greatest integer not exceeding \\( 157.0795 \\) is \\( 157 \\).\n\nThe final answer is \\( \\boxed{157} \\)", "answer": "157", "source": "aops_forum", "problem_type": "Geometry", "uuid": "429b335f-129d-5556-9030-e012a7777eab", "difficulty_band": "medium", "correctness_count": 2}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\nLet $B = (20, 14)$ and $C = (18, 0)$ be two points in the plane. For every line $\\ell$ passing through $B$, we color red the foot of the perpendicular from $C$ to $\\ell$. The set of red points enclose a bounded region of area $\\mathcal{A}$. Find $\\lfloor \\mathcal{A} \\rfloor$ (that is, find the greatest integer not exceeding $\\mathcal A$).\n\n[i]Proposed by Yang Liu[/i]", "candidates": [{"kind": "greedy", "answer": "To find the area enclosed by the set of red points, we need to find the area of the region enclosed by the line $\\ell$ passing through $B = (20, 14)$ and the line $\\ell$ passing through $C = (18, 0)$.\n\nFirst, we need to find the equation of the line passing through $B = (20, 14)$ and $C = (18, 0)$. The slope of this line is given by:\n\\[ m = \\frac{y_2 - y_1}{x_2 - x_1} = \\frac{0 - 14}{18 - 20} = -\\frac{14}{-2} = 7 \\]\n\nThe equation of the line passing through $B$ and $C$ is:\n\\[ y - 14 = 7(x - 20) \\Rightarrow 7x - y - 48 = 0 \\]\n\nNow, we need to find the intersection point of this line and the line passing through $C = (18, 0)$ and $B = (20, 14)$. We can do this by solving the system of equations:\n\\[ 7x - y - 48 = 0 \\quad (1) \\]\n\\[ 7x - y - 48 = 0 \\quad (2) \\]\n\nSubtracting (2) from (1) gives us:\n\\[ 6x = 48 \\Rightarrow x = 8 \\]\n\nSubstituting $x = 8$ into (1) gives us:\n\\[ 7(8) - y - 48 = 0 \\Rightarrow 56 - y = 0 \\Rightarrow y = 56 \\]\n\nNow that we have the coordinates of the intersection point, we can find the area enclosed by the set of red points. This is a triangle with vertices $(20, 14)$, $(18, 0)$, and $(8, 56)$.\n\nThe base of the triangle is the distance between $(20, 14)$ and $(18, 0)$, which is $20 - 18 = 2$. The height of the triangle is the distance from $("}, {"kind": "sampled", "answer": "Assumptions\n\n1. The point $B = (20, 14)$ lies on the line $y = x$ and is on the line $x = 18$.\n2. The point $C = (18, 0)$ lies on the line $y = x$ and lies on the line $x = 18$.\n3. For any line $\\ell$ passing through $B$, we color the foot of the perpendicular from $C$ to $\\ell$ red.\n4. The set of red points enclosed by the bounded region of the half-plane defined by the half-plane defined by the half-plane defined by two red points is the half-plane defined by the line $y = x$.\n\nTo solve this problem, we can break it down into steps.\n\nFirst, we need to find the equation of the line passing through $B = (20, 14)$ and $C = (18, 0)$.\n\nThe slope of this line is:\n\\[ m = \\frac{0 - 14}{18 - 20} = \\frac{-14}{-2} = 7 \\]\n\nThe equation of the line passing through $B$ and $C$ is:\n\\[ y - 14 = 7(x - 20) \\Rightarrow 7x - y - 50 = 0 \\]\n\nNow, we need to find the intersection point of this line and the line $C = (18, 0)$.\n\nTo find the intersection point, we can substitute $y = x$ into the equation of the line passing through $B$ and $C$:\n\\[ 7x - x - 50 = 0 \\Rightarrow 6x = 50 \\Rightarrow x = 25/3 \\]\n\nNow that we have the coordinates of $B$ and $C$, we can find the distance between them. The distance between $B$ and $C$ is:\n\\[ \\text{Distance} = \\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \\sqrt{(25/3 - 18)^2 + (25/3 - 0)^2} = \\sqrt{(-10/3)^"}]} +{"index": 76, "row": {"problem": "Problem: Place the 2004 positive integers $1, 2, \\cdots, 2004$ randomly on a circle. By counting the parity of all adjacent 3 numbers, it is known that there are 600 groups where all 3 numbers are odd, and 500 groups where exactly 2 numbers are odd. How many groups have exactly 1 odd number? How many groups have no odd numbers at all?\n(2004, Shanghai TI Cup High School Mathematics Competition for Grade 10)", "solution": "Solution: These 2004 numbers are arbitrarily placed on a circle, with every 3 adjacent numbers forming a group, making a total of 2004 groups. Each number is part of 3 different groups. Let the number of groups with exactly 1 odd number be $x$, and the number of groups with no odd numbers be $y$. Considering the odd number scenarios, we have\n$$\n600 \\times 3 + 500 \\times 2 + x = 1002 \\times 3 .\n$$\n\nSolving for $x$ gives $x=206$.\nThus, $y=2004-600-500-206=698$.", "answer": "206, 698", "source": "cn_contest", "problem_type": "Combinatorics", "uuid": "0d8c0843-3d6f-5dc2-a66e-a6b3aa236b56", "difficulty_band": "hard", "correctness_count": 1}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\nProblem: Place the 2004 positive integers $1, 2, \\cdots, 2004$ randomly on a circle. By counting the parity of all adjacent 3 numbers, it is known that there are 600 groups where all 3 numbers are odd, and 500 groups where exactly 2 numbers are odd. How many groups have exactly 1 odd number? How many groups have no odd numbers at all?\n(2004, Shanghai TI Cup High School Mathematics Competition for Grade 10)", "candidates": [{"kind": "greedy", "answer": "First, we need to understand the problem. We have 2004 positive integers arranged in a circle. We are asked to find the number of groups with exactly 1 odd number, the number of groups with no odd numbers at all, and the number of groups with no odd numbers at all.\n\nLet's denote the number of groups with 1 odd number as $x$, the number of groups with no odd numbers at all as $y$, and the number of groups with no odd numbers at all as $z$.\n\nWe know that $x + y + z = 2004$.\n\nWe also know that $x$ is the number of groups with 1 odd number, $y$ is the number of groups with no odd numbers at all, and $z$ is the number of groups with no odd numbers at all.\n\nSince each group with 1 odd number is counted in $x$, the number of groups with 1 odd number is $x - z$.\n\nSimilarly, the number of groups with no odd numbers at all is $y - z$, and the number of groups with no odd numbers at all is $z - x$.\n\nSince each group with 1 odd number is counted in $y$, the number of groups with 1 odd number is $y - z$.\n\nSimilarly, the number of groups with no odd numbers at all is $z - x$.\n\nSince each group with 1 odd number is counted in $z$, the number of groups with 1 odd number is $z - x$.\n\nWe know that $x + y + z = 2004$, so we can write $x - z = y - z = z - x = 2004 - (x + y + z)$.\n\nSince $x + y + z = 2004$, we can write $x - z = y - z = z - x = 2004 - (x + y + z)$.\n\nWe can rewrite the equation $x - z = y - z = z - x = 2004 - (x + y + z)$ as $x - z = y - z = z - x = 2004 - (x + y + z)$.\n\nSince $x + y + z = 2004$, we can write $x - z = y - z ="}, {"kind": "sampled", "answer": "This is a combinatorial problem with restrictions, where we have to find the number of groups that have exactly 1 odd number and the number of groups with no odd number.\n\nTo solve this problem, we first need to find the total number of groups. We can do this by finding the number of groups with all 3 odd numbers and subtracting the number of groups with all 3 odd numbers.\n\nThe number of groups with all 3 odd numbers is given by the combination formula:\n\\[ C(n, k) = \\frac{n!}{k!(n-k)!} \\]\n\nFor $n = 2004$ and $k = 2$:\n\\[ C(2004, 2) = \\frac{2004!}{2!(2004-2)!} = \\frac{2004 \\times 2003}{2} = 2004 \\times 1001 \\]\n\nNow, we need to find the number of groups with all 3 odd numbers. To do this, we can find the number of groups with all 3 odd numbers and subtract the number of groups with all 3 odd numbers.\n\nThe number of groups with all 3 odd numbers is given by the combination formula again:\n\\[ C(2004, 3) = \\frac{2004!}{3!(2004-3)!} = \\frac{2004 \\times 2003}{6} = 334604 \\]\n\nThe number of groups with all 3 odd numbers is 334604 - 2004 = 334500\n\nNow, we can find the number of groups with exactly 1 odd number. We can do this by subtracting the number of groups with all 3 odd numbers from the number of groups with all 3 odd numbers.\n\nThe number of groups with exactly 1 odd number is 334500 - 334604 = 100.\n\nTherefore, there are 100 groups that have exactly 1 odd number. The number of groups that have no odd numbers at all is 2004 - 100 = 1904.\n\nSo, the"}]} +{"index": 77, "row": {"problem": "5. (8 points) On the radius $A O$ of a circle with center $O$, a point $M$ is chosen. On one side of $A O$ on the circle, points $B$ and $C$ are chosen such that $\\angle A M B = \\angle O M C = \\alpha$. Find the length of $B C$ if the radius of the circle is $12$, and $\\cos \\alpha = \\frac{3}{4}$?", "solution": "Answer: 18.\n\n## Solution:\n\n![](https://cdn.mathpix.com/cropped/2024_05_06_61450ad4ec176109d5e0g-38.jpg?height=434&width=462&top_left_y=383&top_left_x=794)\n\nConsider point $B_{1}$, which is symmetric to point $B$ with respect to the line $O A$. It also lies on the circle and $\\angle A M B=\\alpha$. Note that points $B_{1}, M, C$ lie on the same straight line, and $\\Delta B B_{1} M$ is isosceles. Therefore, the inscribed angle $\\angle B B_{1} M=90^{\\circ}-\\alpha$, and the central angle $\\angle B O C=180^{\\circ}-2 \\alpha$. $\\triangle B O C$ is isosceles and $\\angle O B C=\\alpha$. We find the base of the isosceles triangle using the formula $B C=2 \\cdot \\cos \\alpha \\cdot B O=2 \\cdot \\frac{3}{4} \\cdot 12=18$", "answer": "18", "source": "olympiads", "problem_type": "Geometry", "uuid": "017bcf16-05db-57e5-8191-4dfbd3b81852", "difficulty_band": "hard", "correctness_count": 1}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\n5. (8 points) On the radius $A O$ of a circle with center $O$, a point $M$ is chosen. On one side of $A O$ on the circle, points $B$ and $C$ are chosen such that $\\angle A M B = \\angle O M C = \\alpha$. Find the length of $B C$ if the radius of the circle is $12$, and $\\cos \\alpha = \\frac{3}{4}$?", "candidates": [{"kind": "greedy", "answer": "(1) Solution:\nLet $M$ be the midpoint of $BC$.\n\nSince $\\angle A M B = \\angle O M C = \\alpha$,\n\n$\\therefore \\angle O B C = 2\\alpha$.\n\n$\\therefore \\angle B C O = 180^\\circ - 2\\alpha$.\n\n$\\therefore \\cos \\alpha = \\frac{1}{2} \\Rightarrow \\alpha = 60^\\circ$.\n\n(2) Solution:\nLet $M$ be the midpoint of $BC$.\n\n$\\therefore \\angle B C O = 180^\\circ - 2\\alpha = 180^\\circ - 120^\\circ = 60^\\circ$.\n\n$\\therefore \\cos \\alpha = \\frac{3}{4}$.\n\n(3) Solution:\nLet $M$ be the midpoint of $BC$.\n\n$\\therefore \\angle B C O = 180^\\circ - 2\\alpha = 180^\\circ - 120^\\circ = 60^\\circ$.\n\n$\\therefore \\cos \\alpha = \\frac{3}{4}$.\n\nFinal answer:\n#### 3"}, {"kind": "sampled", "answer": "1. **Introduce Notations and Initial Conditions**:\n - Let $r$ represent the radius of the original circle.\n - Let $M$ be the point on the radius $OA$ such that $OM = 12$.\n - Let $B$ and $C$ be points on the circle, with $\\angle AOB = \\angle OMC = \\alpha$.\n\n2. **Establish Angles**:\n - Since $M$ is on the radius $OA$, $OM$ is perpendicular to $OA$.\n - Therefore, $\\angle AOC = \\angle AOB = \\alpha$.\n\n3. **Establish Similar Triangles**:\n - Triangles $AOB$ and $OMC$ are similar by AA similarity criterion.\n - This gives us the proportion:\n \\[\n \\frac{OB}{OC} = \\frac{OM}{OC} \\implies \\frac{OB}{12} = \\frac{12}{r}\n \\]\n\n4. **Solve for OB**:\n - To find $OB$, we solve the proportion:\n \\[\n \\frac{OB}{12} = \\frac{12}{r} \\implies OB = 12 \\times \\frac{12}{r} = 144 : r\n \\]\n\n5. **Substitute $\\angle AOB$ in the Expression for $OB$**:\n - Using the angle $\\angle AOB = \\alpha$, we get:\n \\[\n OB = 144 : r = 144 : 12 = 12\n \\]\n\n6. **Relate $ON$ and $OM$**:\n - From the similarity and proportions, we have:\n \\[\n ON = \\frac{OM \\cdot OB}{OB} = \\frac{12 \\cdot 12}{144 : r} = \\frac{144}{144 : r} = \\frac{144r}{144} = 1\n \\]\n\n7. **Calculate $OB$**:\n - Since $OB = 1$, we find:\n \\[\n OB = \\boxed{12}\n \\]\n\nFinal answer:\n#### 12"}]} +{"index": 78, "row": {"problem": "One of the receipts for a math tournament showed that $72$ identical trophies were purchased for $\\$$-$99.9$-, where the first and last digits were illegible. How much did each trophy cost?", "solution": "1. **Understanding the Problem:**\n We need to determine the cost of each trophy given that 72 identical trophies were purchased for a total price of $\\$ -99.9-$, where the first and last digits are illegible.\n\n2. **Convert the Total Price to Cents:**\n Let the total price be $x99.9y$ dollars, where $x$ and $y$ are the unknown digits. Converting this to cents, we get:\n \\[\n 100 \\times (x99.9y) = 1000x + 9990 + y\n \\]\n Therefore, the total price in cents is:\n \\[\n 1000x + 9990 + y\n \\]\n\n3. **Divisibility by 72:**\n The total price in cents must be divisible by 72. Since 72 is the product of 8 and 9, the total price must be divisible by both 8 and 9.\n\n4. **Divisibility by 8:**\n For a number to be divisible by 8, its last three digits must be divisible by 8. The last three digits of our number are $99y$. We need to find $y$ such that $99y$ is divisible by 8.\n \\[\n 990 + y \\equiv 0 \\pmod{8}\n \\]\n Testing values for $y$:\n \\[\n 990 \\equiv 6 \\pmod{8} \\implies 6 + y \\equiv 0 \\pmod{8} \\implies y \\equiv 2 \\pmod{8}\n \\]\n Therefore, $y = 2$.\n\n5. **Divisibility by 9:**\n For a number to be divisible by 9, the sum of its digits must be divisible by 9. The sum of the digits of $x99.92$ is:\n \\[\n x + 9 + 9 + 9 + 2 = x + 29\n \\]\n We need $x + 29$ to be divisible by 9. Testing values for $x$:\n \\[\n x + 29 \\equiv 0 \\pmod{9}\n \\]\n \\[\n x \\equiv -29 \\pmod{9} \\implies x \\equiv 7 \\pmod{9}\n \\]\n Therefore, $x = 7$.\n\n6. **Total Price:**\n The total price is $799.92$ dollars.\n\n7. **Cost per Trophy:**\n Dividing the total price by the number of trophies:\n \\[\n \\frac{799.92}{72} = 11.11\n \\]\n Therefore, the cost of each trophy is $\\$11.11$.\n\nThe final answer is $\\boxed{11.11}$.", "answer": "11.11", "source": "aops_forum", "problem_type": "Algebra", "uuid": "9d20d7e6-3677-5aa5-ad27-a43c8843bc6f", "difficulty_band": "medium", "correctness_count": 2}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\nOne of the receipts for a math tournament showed that $72$ identical trophies were purchased for $\\$$-$99.9$-, where the first and last digits were illegible. How much did each trophy cost?", "candidates": [{"kind": "greedy", "answer": "The first and last digits are 72, so the total cost is 72 - 99.9 = 2.1.\n\nThe answer is \\boxed{2.1}."}, {"kind": "sampled", "answer": "The first and last digits of the four-digit number are 72. The total cost is 99-72=27."}]} +{"index": 79, "row": {"problem": "It is known that the equation$ |x - 1| + |x - 2| +... + |x - 2001| = a$ has exactly one solution. Find $a$.", "solution": "1. **Identify the nature of the function:**\n The given function is \\( f(x) = |x - 1| + |x - 2| + \\cdots + |x - 2001| \\). This function is a sum of absolute values, which is piecewise linear and changes slope at each of the points \\( x = 1, 2, \\ldots, 2001 \\).\n\n2. **Determine the median:**\n The function \\( f(x) \\) achieves its minimum value when \\( x \\) is the median of the set \\(\\{1, 2, \\ldots, 2001\\}\\). Since there are 2001 points, the median is the middle value, which is \\( x = 1001 \\).\n\n3. **Calculate the value of the function at the median:**\n To find \\( a \\), we need to evaluate \\( f(1001) \\):\n \\[\n f(1001) = |1001 - 1| + |1001 - 2| + \\cdots + |1001 - 2001|\n \\]\n This can be split into two sums:\n \\[\n f(1001) = \\sum_{i=1}^{1000} (1001 - i) + \\sum_{i=1002}^{2001} (i - 1001)\n \\]\n\n4. **Evaluate the first sum:**\n \\[\n \\sum_{i=1}^{1000} (1001 - i) = \\sum_{i=1}^{1000} 1001 - \\sum_{i=1}^{1000} i = 1000 \\cdot 1001 - \\frac{1000 \\cdot 1001}{2} = 1000 \\cdot 1001 - 500 \\cdot 1001 = 500 \\cdot 1001\n \\]\n\n5. **Evaluate the second sum:**\n \\[\n \\sum_{i=1002}^{2001} (i - 1001) = \\sum_{i=1}^{1000} i = \\frac{1000 \\cdot 1001}{2} = 500 \\cdot 1001\n \\]\n\n6. **Combine the results:**\n \\[\n f(1001) = 500 \\cdot 1001 + 500 \\cdot 1001 = 1000 \\cdot 1001 = 1001000\n \\]\n\nConclusion:\n\\[\na = 1001000\n\\]\n\nThe final answer is \\(\\boxed{1001000}\\).", "answer": "1001000", "source": "aops_forum", "problem_type": "Other", "uuid": "4ac75d6f-6137-5e60-b43e-fb0babe826df", "difficulty_band": "medium", "correctness_count": 2}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\nIt is known that the equation$ |x - 1| + |x - 2| +... + |x - 2001| = a$ has exactly one solution. Find $a$.", "candidates": [{"kind": "greedy", "answer": "The equation $|x - 1| + |x - 2| +... + |x - 2001| = a$ has exactly one solution if and only if the equation $|x - 1| + |x - 2| +... + |x - 2001| = 0$ has exactly one solution.\n\nThe equation $|x - 1| + |x - 2| +... + |x - 2001| = 0$ is equivalent to $|x - 1| = -(|x - 2| +... + |x - 2001|)$.\n\nThis means that $|x - 1| = -|x - 2| -... - |x - 2001|$.\n\nSince $|x - 1|$ is an even function, we have $|x - 1| = |x - 2| +... + |x - 2001|$.\n\nTherefore, $|x - 1| = -|x - 2| -... - |x - 2001|$.\n\nThis means that $|x - 1| = -|x - 2| -... - |x - 2001|$.\n\nSince $|x - 1|$ is an even function, we have $|x - 1| = |x - 2| +... + |x - 2001|$.\n\nTherefore, $|x - 1| = -|x - 2| -... - |x - 2001|$.\n\nThis means that $|x - 1| = -|x - 2| -... - |x - 2001|$.\n\nSince $|x - 1|$ is an even function, we have $|x - 1| = |x - 2| +... + |x - 2001|$.\n\nTherefore, $|x - 1| = -|x - 2| -... - |x - 2001|$.\n\nThis means that $|x - 1| = -|x - 2| -... - |x - 2001|$.\n\nSince $|x"}, {"kind": "sampled", "answer": "The equation $ |x - 1| + |x - 2| +... + |x - 2001| = a$ has exactly one solution when the number of terms in the sum is 1.\n\nTherefore, we must have $ |x - 1| + |x - 2| +... + |x - 2001| = 2002 - a$.\n\nThe only way this equation can have a unique solution is when the sum includes the term $2002 - a$.\n\nTherefore, $ a = 2002 - (|x - 1| + |x - 2| +... + |x - 2001|) = 2002 - (2002 - |x - 1| + |x - 2| +... + |x - 2001|) = 2002 - (2002 - (2002 - a)) = a - a = 2002 - 2a = 2002 - (2002 - 2a - a) = 2002 - 2a - 2a = 2002 - 4a = 2002 - 4(2002 - a - a) = 2002 - 4a - 2002 + 4a = 4a - 2a = 2a = 2002 - 2a \\Rightarrow a = 2002 - 2a \\Rightarrow 3a = 2002 \\Rightarrow a = \\boxed{2002 : 3} \\]"}]} +{"index": 80, "row": {"problem": "Example 5. Expand the function\n\n$$\nf(z)=\\frac{z}{z^{2}-2 z-3}\n$$\n\ninto a Taylor series in the neighborhood of the point $z_{0}=0$ using expansion (12), and find the radius of convergence of the series.", "solution": "Solution. Let's decompose the given function into partial fractions:\n\n$$\n\\frac{z}{z^{2}-2 z-3}=\\frac{1}{4} \\frac{1}{z+1}-\\frac{3}{4} \\frac{1}{z-3}\n$$\n\nTransform the right-hand side as follows:\n\n$$\nf(z)=\\frac{1}{4} \\frac{1}{1+z}-\\frac{1}{4} \\frac{1}{1-\\frac{2}{3}}\n$$\n\nUsing the expansion (12) of the function $\\frac{1}{1+z}$, we get\n\n$$\n\\begin{aligned}\nf(z) & =\\frac{1}{4}\\left(1-z+z^{2}-z^{3}+\\ldots\\right)-\\frac{1}{4}\\left(1+\\frac{z}{3}+\\frac{z^{2}}{9}+\\ldots\\right)= \\\\\n& =\\frac{1}{4}\\left(-\\frac{4}{3} z+\\frac{8}{9} z^{2}-\\frac{28}{27} z^{3}+\\ldots\\right)=-\\frac{z}{3}+\\frac{2}{3^{2}} z^{2}-\\frac{7}{3^{3}} z^{3}+\\ldots\n\\end{aligned}\n$$\n\nThe nearest singular point to the point $z_{0}=0$ of the given function is the point $z=-1$. Therefore, the radius of convergence of the obtained series is $R=1$.", "answer": "1", "source": "olympiads", "problem_type": "Algebra", "uuid": "632d4671-91dd-5893-a0e2-fcd043c434af", "difficulty_band": "medium", "correctness_count": 2}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\nExample 5. Expand the function\n\n$$\nf(z)=\\frac{z}{z^{2}-2 z-3}\n$$\n\ninto a Taylor series in the neighborhood of the point $z_{0}=0$ using expansion (12), and find the radius of convergence of the series.", "candidates": [{"kind": "greedy", "answer": "To expand the function $f(z)=\\frac{z}{z^{2}-2 z-3}$ into a Taylor series in the neighborhood of the point $z_{0}=0$, we can use the method of Taylor series expansion.\n\nThe Taylor series expansion of $f(z)$ is given by:\n\n$$\nf(z) = \\sum_{n=0}^{\\infty} \\frac{z^{n+1}}{(z-n)^{2}+n(z-n)^{2}}\n$$\n\nWe want to find the radius of convergence of this series. To do this, we can use the fact that the series converges if and only if the absolute value of the derivative of the series is less than 1.\n\nFirst, we need to find the derivative of $f(z)$:\n\n$$\nf'(z) = \\frac{d}{dz} \\left( \\frac{z}{z^{2}-2 z-3} \\right) = \\frac{(z^{2}-2 z-3)(1) - z(2z-3)}{(z^{2}-2 z-3)^2}\n$$\n\nNow, we can simplify the expression for $f'(z)$:\n\n$$\nf'(z) = \\frac{z^{2}-2 z-3 - 2z^{2} + 6z + 3z - 9}{(z^{2}-2 z-3)^2} = \\frac{-z^{2} + 4z + 6}{(z^{2}-2 z-3)^2}\n$$\n\nNext, we need to find the radius of convergence of this expression. To do this, we can use the fact that the series converges if and only if the absolute value of the derivative of the series is less than 1.\n\nThe derivative of $f'(z)$ is:\n\n$$\nf'(z) = \\frac{d}{dz} \\left( \\frac{-z^{2} + 4z + 6}{(z^{2}-2 z-3)^2} \\right) = \\frac{(z^{2}-2 z-3)(2z-6) - z(2z-3)}{(z^{2}-2 z-3)^2}\n$$\n\nNow"}, {"kind": "sampled", "answer": "The Taylor series expansion of $f(z) = \\frac{z}{z^{2} - 2 z - 3}$ is given by the derivatives of $f(z)$ at $z = 0$ and $z = 2$.\n\nTo find the radius of convergence of the Taylor series, we need to find the radius of convergence of the geometric series of derivatives at $z = 0$ and $z = 2$.\n\nThe radius of convergence is given by:\n\n$$R = \\frac{|\\frac{d}{dz}f(z) - \\frac{d}{dz}f'(z)}{\\frac{d}{dz}f(z) + \\frac{d}{dz}f'(z)}$$\n\nwhere $d$ is the first derivative of $f(z)$.\n\nSince $f'(z) = 2z + 2$, we can substitute this into the expression for $R$:\n\n$$R = \\frac{|\\frac{d}{dz}f(z) - 2z - 2}{2z + 2 + 2}$$\n\nNow, we can expand the numerator of $R$ and simplify it.\n\nAfter simplification, we get:\n\n$$R = \\frac{|z| - 2}{4z + 4}$$\n\nThe radius of convergence is:\n\n$$R = \\frac{|\\frac{|z| - 2}{4z + 4} - 2|}{4}$$\n\nSince the radius of convergence is given by the limit as $z$ goes to $0$, we can find the limit as $z$ goes to $0$:\n\n$$R = \\lim_{z \\to 0} \\frac{|z| - 2}{4}$$\n\nThe limit as $z$ goes to $0$ is:\n\n$$\\lim_{z \\to 0} \\frac{|z| - 2}{4} = \\frac{0 - 2}{4} = -\\frac{1}{2}$$\n\nTherefore, the radius of convergence of the Taylor series is:\n\n$$R = \\boxed{-\\frac{1}{2}}$$"}]} +{"index": 81, "row": {"problem": "Problem 5. An electric kettle heats water from room temperature $T_{0}=20^{\\circ} \\mathrm{C}$ to $T_{m}=100^{\\circ} \\mathrm{C}$ in $t=10$ minutes. How long will it take $t_{1}$ for all the water to boil away if the kettle is not turned off and the automatic shut-off system is faulty? The specific heat capacity of water $c=4200$ J/kg $\\cdot$ K. The specific latent heat of vaporization of water $L=2.3$ MJ/kg. Round the answer to the nearest whole number of minutes.", "solution": "Solution. The power $P$ of the kettle is fixed and equal to $P=Q / t$. From the heat transfer law $Q=c m\\left(T_{m}-T_{0}\\right)$ we get $P t=c m\\left(T_{m}-T_{0}\\right)$.\n\nTo evaporate the water, the amount of heat required is $Q_{1}=L m \\Rightarrow P t_{1}=L m$.\n\nBy comparing these relations, we obtain $\\frac{t_{1}}{t}=\\frac{L m}{c m\\left(T_{m}-T_{0}\\right)}=\\frac{L}{c\\left(T_{m}-T_{0}\\right)} \\Rightarrow$\n\n$$\n\\frac{t_{1}}{t}=\\frac{2.3 \\cdot 10^{6}}{4200 \\cdot 80}=\\frac{2300}{336} \\approx 6.845\n$$\n\nThus, the water will boil away in $t_{1}=6.845 \\cdot t=68.45$ minutes.\n\nAnswer: 68 minutes.", "answer": "68", "source": "olympiads", "problem_type": "Algebra", "uuid": "25820c70-8e88-59c2-bbcd-8d4a37372400", "difficulty_band": "hard", "correctness_count": 1}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\nProblem 5. An electric kettle heats water from room temperature $T_{0}=20^{\\circ} \\mathrm{C}$ to $T_{m}=100^{\\circ} \\mathrm{C}$ in $t=10$ minutes. How long will it take $t_{1}$ for all the water to boil away if the kettle is not turned off and the automatic shut-off system is faulty? The specific heat capacity of water $c=4200$ J/kg $\\cdot$ K. The specific latent heat of vaporization of water $L=2.3$ MJ/kg. Round the answer to the nearest whole number of minutes.", "candidates": [{"kind": "greedy", "answer": "The specific heat of gas is $L=2.3$ J/kg.\n\nThe specific heat of gas is $L=2.3$ J/kg.\n\nThe specific heat of gas is $L=2.3$ J/kg.\n\nThe specific heat of gas is $L=2.3$ J/kg.\n\nThe specific heat of gas is $L=2.3$ J/kg.\n\nThe specific heat of gas is $L=2.3$ J/kg.\n\nThe specific heat of gas is $L=2.3$ J/kg.\n\nThe specific heat of gas is $L=2.3$ J/kg.\n\nThe specific heat of gas is $L=2.3$ J/kg.\n\nThe specific heat of gas is $L=2.3$ J/kg.\n\nThe specific heat of gas is $L=2.3$ J/kg.\n\nThe specific heat of gas is $L=2.3$ J/kg.\n\nThe specific heat of gas is $L=2.3$ J/kg.\n\nThe specific heat of gas is $L=2.3$ J/kg.\n\nThe specific heat of gas is $L=2.3$ J/kg.\n\nThe specific heat of gas is $L=2.3$ J/kg.\n\nThe specific heat of gas is $L=2.3$ J/kg.\n\nThe specific heat of gas is $L=2.3$ J/kg.\n\nThe specific heat of gas is $L=2.3$ J/kg.\n\nThe specific heat of gas is $L=2.3$ J/kg.\n\nThe specific heat of gas is $L=2.3$ J/kg.\n\nThe specific heat of gas is $L=2.3$ J/kg.\n\nThe specific heat of gas is $L=2.3$ J/kg.\n\nThe specific heat of gas is $L=2.3$ J/kg.\n\nThe specific heat of gas is $L=2.3$ J/kg.\n\nThe specific heat of gas is $L=2.3$ J/kg.\n\nThe specific heat of gas is $L=2.3$ J/kg."}, {"kind": "sampled", "answer": "The specific heat capacity of water $c=4200$ J/kg $\\cdot$ K.\n\nThe specific Masked heat capacity of liquid $L=2.3$ inhib/kg.\n\nIn this problem, the heat capacity of liquid $L$ is 2.3pen(M\\,T) = 2.3(4200\\,K) = 9100\\,J/kg.\n\nTherefore, the answer is:\n\\[ \\boxed{9100} \\]"}]} +{"index": 82, "row": {"problem": "5. As shown in the figure, quadrilaterals $A B C D$ and $C E F G$ are two squares. Given that $\\frac{S_{\\triangle A B I}}{S_{\\triangle E F I}}=27$, find $\\frac{A B}{E F}$.", "solution": "【Analysis】Let $A B=x, E F=y$,\n$$\n\\begin{array}{l}\n\\frac{A I}{I E}=\\frac{S_{\\triangle A B F}}{S_{\\triangle B E F}}=\\frac{\\frac{1}{2} \\times x \\times(x+y)}{\\frac{1}{2} \\times y \\times y}=\\frac{x(x+y)}{y^{2}} \\\\\n\\frac{I F}{I B}=\\frac{S_{\\triangle A E F}}{S_{\\triangle A B E}}=\\frac{\\frac{1}{2} \\times y \\times(x+y)}{\\frac{1}{2} \\times x \\times x}=\\frac{x(x+y)}{x^{2}}\n\\end{array}\n$$\n$S_{\\triangle A B I}$ and $S_{\\triangle E F I}$ share an angle (bird head model)\n$$\n\\frac{S_{\\triangle A B I}}{S_{\\triangle E F I}}=\\frac{A I \\times B I}{I E \\times I F}=\\frac{x(x+y)}{y^{2}} \\times \\frac{x^{2}}{y(x+y)}=\\frac{x^{3}}{y^{3}}=27\n$$\n\nSo $\\frac{A B}{E F}=\\frac{x}{y}=3$.\n【Answer】3", "answer": "3", "source": "olympiads", "problem_type": "Geometry", "uuid": "f5dbbed1-66ec-5942-8594-f6de0862d0b0", "difficulty_band": "hard", "correctness_count": 1}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\n5. As shown in the figure, quadrilaterals $A B C D$ and $C E F G$ are two squares. Given that $\\frac{S_{\\triangle A B I}}{S_{\\triangle E F I}}=27$, find $\\frac{A B}{E F}$.", "candidates": [{"kind": "greedy", "answer": "1. Let's denote the side length of the square $A B C D$ as $s$.\n2. The area of the square $A B C D$ is $s^2$.\n3. The area of the square $C E F G$ is $s^2$.\n4. The area of the square $A B I$ is $s^2$.\n5. The area of the square $E F I$ is $s^2$.\n6. The area of the square $A B I$ is $s^2$.\n7. The area of the square $E F G$ is $s^2$.\n8. The area of the square $A B C D$ is $s^2$.\n9. The area of the square $C E F G$ is $s^2$.\n10. The area of the square $A B I$ is $s^2$.\n11. The area of the square $E F I$ is $s^2$.\n12. The area of the square $A B C D$ is $s^2$.\n13. The area of the square $C E F G$ is $s^2$.\n14. The area of the square $A B I$ is $s^2$.\n15. The area of the square $E F I$ is $s^2$.\n16. The area of the square $A B C D$ is $s^2$.\n17. The area of the square $C E F G$ is $s^2$.\n18. The area of the square $A B I$ is $s^2$.\n19. The area of the square $E F I$ is $s^2$.\n20. The area of the square $A B C D$ is $s^2$.\n21. The area of the square $C E F G$ is $s^2$.\n22. The area of the square $A B I$ is $s^2$.\n23. The area of the square $E F I$ is $s^2$.\n24. The area of the square $A B C D$ is $s^2$.\n25. The area of the square $C E F G$ is $s^2$.\n26. The area of"}, {"kind": "sampled", "answer": "1. We start by noting that the ratio of the areas of the two squares is given by:\n \\[\n \\frac{S_{\\triangle I A B}}{S_{\\triangle E F I}} = \\frac{S_{\\triangle I A B}}{\\frac{S_{\\triangle E F I}}{2}} = \\frac{S_{\\triangle I A B}}{\\frac{2 S_{\\triangle E F I}}{S_{\\triangle E F I}}} = \\frac{S_{\\triangle I A B}}{\\frac{2 S_{\\triangle E F I}}{S_{\\triangle E F I}}} = \\frac{S_{\\triangle I A B}}{\\frac{2}{\\frac{S_{\\triangle E F I}}{S_{\\triangle E F I}}}} = S_{\\triangle I A B} \\times \\frac{S_{\\triangle E F I}}{2}\n \\]\n\n2. Next, we note that the ratio of the areas of the two squares is given by:\n \\[\n \\frac{S_{\\triangle E F I}}{S_{\\triangle I A B}} = \\frac{S_{\\triangle E F I}}{S_{\\triangle I A B} \\frac{S_{\\triangle E F I}}{2}} = \\frac{S_{\\triangle E F I} \\times \\frac{S_{\\triangle E F I}}{2}}{S_{\\triangle I A B} \\times S_{\\triangle E F I} \\frac{S_{\\triangle E F I}}{2}} = \\frac{S_{\\triangle E F I}}{S_{\\triangle I A B} \\frac{S_{\\triangle E F I}}{2}} = \\frac{S_{\\triangle E F I}}{S_{\\triangle I A B} \\frac{S_{\\triangle E F I}}{2}} = \\frac{S_{\\triangle E F I}}{S_{\\triangle I A B} S_{\\triangle E F I} \\times 2} = \\frac{S_{\\triangle E F I}}{S_{\\triangle I A B} S_{\\triangle E F"}]} +{"index": 83, "row": {"problem": "Triangle $ABC$ is inscribed in a unit circle $\\omega$. Let $H$ be its orthocenter and $D$ be the foot of the perpendicular from $A$ to $BC$. Let $\\triangle XY Z$ be the triangle formed by drawing the tangents to $\\omega$ at $A, B, C$. If $\\overline{AH} = \\overline{HD}$ and the side lengths of $\\triangle XY Z$ form an arithmetic sequence, the area of $\\triangle ABC$ can be expressed in the form $\\tfrac{p}{q}$ for relatively prime positive integers $p, q$. What is $p + q$?", "solution": "1. **Understanding the Problem:**\n - We have a triangle \\( \\triangle ABC \\) inscribed in a unit circle \\( \\omega \\).\n - \\( H \\) is the orthocenter of \\( \\triangle ABC \\).\n - \\( D \\) is the foot of the perpendicular from \\( A \\) to \\( BC \\).\n - \\( \\triangle XYZ \\) is formed by drawing tangents to \\( \\omega \\) at \\( A, B, C \\).\n - Given \\( \\overline{AH} = \\overline{HD} \\) and the side lengths of \\( \\triangle XYZ \\) form an arithmetic sequence.\n - We need to find the area of \\( \\triangle ABC \\) in the form \\( \\frac{p}{q} \\) and determine \\( p + q \\).\n\n2. **Orthocenter Condition:**\n - Given \\( \\overline{AH} = \\overline{HD} \\), we know that \\( H \\) lies on the nine-point circle of \\( \\triangle ABC \\), which is also the circumcircle of the medial triangle.\n - Reflecting \\( H \\) over \\( D \\) to \\( H' \\), we know \\( H' \\) lies on \\( \\omega \\).\n\n3. **Using Trigonometry and Power of a Point:**\n - Let \\( h \\) be the height from \\( A \\) to \\( BC \\).\n - Using the power of a point:\n \\[\n BD \\cdot DC = AD \\cdot DH' \\implies h \\cot B \\cdot h \\cot C = h \\cdot \\frac{h}{2}\n \\]\n \\[\n \\implies \\tan B \\cdot \\tan C = 2\n \\]\n\n4. **Tangents and Inradius:**\n - Let \\( I \\) be the incenter of \\( \\triangle XYZ \\).\n - Noting that \\( \\angle ZIA = \\angle BCA \\) and \\( \\angle YIA = \\angle ABC \\):\n \\[\n \\tan B \\cdot \\tan C = \\frac{AZ}{1} \\cdot \\frac{AY}{1} = 2\n \\]\n\n5. **Side Lengths of \\( \\triangle XYZ \\):**\n - Let \\( XB = XC = x \\), \\( YA = YC = y \\), and \\( ZA = ZB = z \\).\n - Given \\( 2yz = 2 \\), we have \\( yz = 1 \\).\n - The inradius of \\( \\triangle XYZ \\) is \\( \\sqrt{\\frac{xyz}{x+y+z}} \\), which equals 1 (since the circumradius of \\( \\triangle ABC \\) is 1):\n \\[\n x + y + z = xyz \\implies x + y + z = 2x \\implies x = y + z\n \\]\n\n6. **Arithmetic Sequence Condition:**\n - Since the side lengths form an arithmetic sequence, we can assume \\( x = y + z \\).\n - Let \\( 2z = x + y \\):\n \\[\n 2z = x + y \\implies 2z = 2y + z \\implies z = 2y \\implies z = 2, y = 1, x = 3\n \\]\n\n7. **Area Calculation:**\n - The side lengths \\( x = 3 \\), \\( y = 1 \\), \\( z = 2 \\) form a \\( 3-4-5 \\) triangle.\n - The area of \\( \\triangle ABC \\) can be calculated by subtracting the outer three triangles from the total area:\n \\[\n \\text{Area of } \\triangle ABC = \\frac{6}{5}\n \\]\n\nThe final answer is \\( \\boxed{11} \\)", "answer": "11", "source": "aops_forum", "problem_type": "Geometry", "uuid": "194de747-ae27-5347-817c-5fac137090a7", "difficulty_band": "hard", "correctness_count": 1}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\nTriangle $ABC$ is inscribed in a unit circle $\\omega$. Let $H$ be its orthocenter and $D$ be the foot of the perpendicular from $A$ to $BC$. Let $\\triangle XY Z$ be the triangle formed by drawing the tangents to $\\omega$ at $A, B, C$. If $\\overline{AH} = \\overline{HD}$ and the side lengths of $\\triangle XY Z$ form an arithmetic sequence, the area of $\\triangle ABC$ can be expressed in the form $\\tfrac{p}{q}$ for relatively prime positive integers $p, q$. What is $p + q$?", "candidates": [{"kind": "greedy", "answer": "Since $\\triangle XY Z$ is formed by drawing the tangents to $\\omega$ at $A, B, C$, we can see that $\\triangle XY Z$ is similar to $\\triangle ABC$ with a ratio of $\\frac{1}{2}$.\n\nLet $r$ be the radius of the unit circle. Then $r = 1$.\n\nSince $\\triangle XY Z$ is similar to $\\triangle ABC$, we have:\n\\[ \\frac{r}{1} = \\frac{1}{2} \\Rightarrow r = \\frac{1}{2} \\]\n\nThe area of $\\triangle ABC$ is:\n\\[ \\frac{1}{2} \\cdot 1 \\cdot 1 = \\frac{1}{2} \\]\n\nThe area of $\\triangle XY Z$ is:\n\\[ \\frac{1}{2} \\cdot \\frac{1}{2} \\cdot \\frac{1}{2} = \\frac{1}{8} \\]\n\nThe area of $\\triangle ABC$ is:\n\\[ \\frac{1}{2} \\cdot \\frac{1}{2} = \\frac{1}{4} \\]\n\nThe area of $\\triangle XYZ$ is:\n\\[ \\frac{1}{2} \\cdot \\frac{1}{2} \\cdot \\frac{1}{2} = \\frac{1}{8} \\]\n\nThe area of $\\triangle XYZ$ is:\n\\[ \\frac{1}{2} \\cdot \\frac{1}{2} = \\frac{1}{4} \\]\n\nThe area of $\\triangle XYZ$ is:\n\\[ \\frac{1}{2} \\cdot \\frac{1}{2} = \\frac{1}{4} \\]\n\nThe area of $\\triangle XYZ$ is:\n\\[ \\frac{1}{2} \\cdot \\frac{1}{2} = \\frac{1}{4} \\]\n\nThe area of $\\triangle XYZ$ is:\n\\[ \\frac{1}{2} \\cdot \\frac{1}{2} = \\frac{1}{4} \\]\n\nThe area of $\\triangle XYZ$ is:\n\\[ \\frac{1}{2} \\cdot \\frac{1}{2} = \\frac{1}{"}, {"kind": "sampled", "answer": "Since $\\triangle XYZ$ is formed by drawing the tangents to $\\omega$ at $A$, $B$, and $C$, it is an isosceles triangle, so the angles opposite the equal sides are equal.\n\nLet $O$ be the center of the circle, and $A'$ be the midpoint of $BC$. Then $OA' = OB' = OC' = \\frac{1}{2}BC$.\n\nSince $\\triangle A'BZ$ is isosceles, we have $\\angle BZO = \\angle BKA' = \\angle ZΔ = \\angle DAB$.\n\nSince $\\angle DAB = \\angle DBC$, we have $\\angle DBC = \\angle BZO = \\angle ZΔ$.\n\nLet $M$ be the midpoint of $BC$. Then $BM = \\frac{1}{2}BC$.\n\nSince $\\triangle XYZ$ is isosceles, we have $XY = XZ$.\n\nNow, notice that $\\angle XZD = \\angle A'DZ$ and $\\angle XDZ = \\angle BZH$.\n\nSince $\\angle XDZ = \\angle BZH$, we have that $\\triangle XDZ$ is similar to $\\triangle XZH$.\n\nTherefore, we have that $\\frac{XD}{XZ} = \\frac{AH}{HD}$.\n\nSince $H$ is the orthocenter of $\\triangle ABC$, we have that $H$ is the circumcenter of $\\triangle ABC$.\n\nTherefore, we have that $\\frac{XD}{XZ} = \\frac{AH}{HD}$.\n\nSince $AH = HD$, we have that $\\frac{XD}{XZ} = \\frac{XH}{HZ}$.\n\nSince $H$ is the circumcenter of $\\triangle ABC$, we have that $H$ is the circumcenter of $\\triangle ABC$.\n\nTherefore, we have that $\\frac{XD}{XZ} = \\frac{AH}{HD}$.\n\nSince $XY = XZ$, we have that $\\frac{XD}{XZ} = \\frac{XH}{HZ}$.\n\nSince $X$ is the midpoint of $BC$ and $H$ is the circumcenter of $\\triangle"}]} +{"index": 84, "row": {"problem": "Let $S_n$ be the sum of the reciprocals of the non-zero digits of the integers from 1 to $10^n$ inclusive. Find the smallest positive integer $n$ for which $S_n$ is an integer.", "solution": "1. **Understanding the problem**: We need to find the smallest positive integer \\( n \\) for which the sum of the reciprocals of the non-zero digits of the integers from 1 to \\( 10^n \\) inclusive, denoted as \\( S_n \\), is an integer.\n\n2. **Excluding \\( 10^n \\)**: The number \\( 10^n \\) has only one non-zero digit, which is 1. The reciprocal of 1 is 1, which is an integer. Therefore, \\( 10^n \\) does not affect the integrality of \\( S_n \\).\n\n3. **Including 0**: The number 0 is not included in the range from 1 to \\( 10^n \\), so we do not need to consider it.\n\n4. **Symmetry and Grouping**: Every number from 1 to \\( 10^n - 1 \\) can be expressed as an \\( n \\)-digit number with leading zeros allowed. We can group numbers based on their digits. For example, the numbers 0113457789, 1224568890, and 2335679901 would be in the same group if we consider their digits cyclically.\n\n5. **Number of Groups**: Each group has 10 numbers because there are 10 possible digits (0 through 9). Therefore, there are \\( 10^{n-1} \\) groups.\n\n6. **Sum of Reciprocals in Each Group**: For each group, each place value contributes the sum of the reciprocals of the non-zero digits:\n \\[\n \\frac{1}{1} + \\frac{1}{2} + \\cdots + \\frac{1}{8} + \\frac{1}{9}\n \\]\n This sum is:\n \\[\n H_9 - 1 = \\left(1 + \\frac{1}{2} + \\frac{1}{3} + \\cdots + \\frac{1}{9}\\right) - 1\n \\]\n where \\( H_9 \\) is the 9th harmonic number.\n\n7. **Total Sum \\( S_n \\)**: The total sum \\( S_n \\) is given by:\n \\[\n S_n = n \\cdot 10^{n-1} \\cdot \\left(\\frac{1}{1} + \\frac{1}{2} + \\cdots + \\frac{1}{8} + \\frac{1}{9}\\right)\n \\]\n Simplifying, we get:\n \\[\n S_n = n \\cdot 10^{n-1} \\cdot (H_9 - 1)\n \\]\n\n8. **Finding \\( n \\)**: We need \\( S_n \\) to be an integer. Therefore, \\( n \\cdot 10^{n-1} \\cdot (H_9 - 1) \\) must be an integer. The sum \\( H_9 - 1 \\) is:\n \\[\n H_9 - 1 = \\left(1 + \\frac{1}{2} + \\frac{1}{3} + \\cdots + \\frac{1}{9}\\right) - 1\n \\]\n This sum is not an integer, so we need to find the smallest \\( n \\) such that \\( n \\cdot 10^{n-1} \\) cancels out the fractional part of \\( H_9 - 1 \\).\n\n9. **Calculating \\( H_9 \\)**:\n \\[\n H_9 = 1 + \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{4} + \\frac{1}{5} + \\frac{1}{6} + \\frac{1}{7} + \\frac{1}{8} + \\frac{1}{9}\n \\]\n \\[\n H_9 \\approx 2.828968253968254\n \\]\n \\[\n H_9 - 1 \\approx 1.828968253968254\n \\]\n\n10. **Finding the Smallest \\( n \\)**: We need \\( n \\cdot 10^{n-1} \\) to be a multiple of the denominator of the fractional part of \\( H_9 - 1 \\). The fractional part is approximately 0.828968253968254, which suggests a denominator of 63 (since 63 is the smallest number that makes the fractional part an integer).\n\n11. **Conclusion**: The smallest \\( n \\) that makes \\( S_n \\) an integer is \\( n = 63 \\).\n\nThe final answer is \\( \\boxed{63} \\)", "answer": "63", "source": "aops_forum", "problem_type": "Number Theory", "uuid": "2e0bb6d0-0e35-5b59-a059-41124a8b8e15", "difficulty_band": "hard", "correctness_count": 1}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\nLet $S_n$ be the sum of the reciprocals of the non-zero digits of the integers from 1 to $10^n$ inclusive. Find the smallest positive integer $n$ for which $S_n$ is an integer.", "candidates": [{"kind": "greedy", "answer": "To find the smallest positive integer $n$ for which $S_n$ is an integer, we need to consider the sum of the reciprocals of the non-zero digits of the integers from 1 to $10^n$.\n\nThe non-zero digits of the integers from 1 to $10^n$ are 1, 2, 3,..., $9 \\cdot 10^{n-1}$.\n\nThe sum of the reciprocals of these digits is:\n\\[ S_n = \\frac{1}{1} + \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{4} + \\frac{1}{5} + \\frac{1}{6} + \\frac{1}{7} + \\frac{1}{8} + \\frac{1}{9} + \\frac{1}{10} + \\frac{1}{11} + \\frac{1}{12} + \\frac{1}{13} + \\frac{1}{14} + \\frac{1}{15} + \\frac{1}{16} + \\frac{1}{17} + \\frac{1}{18} + \\frac{1}{19} + \\frac{1}{20} + \\frac{1}{21} + \\frac{1}{22} + \\frac{1}{23} + \\frac{1}{24} + \\frac{1}{25} + \\frac{1}{26} + \\frac{1}{27} + \\frac{1}{28} + \\frac{1}{29} + \\frac{1}{30} + \\frac{1}{31} + \\frac{1}{32} + \\frac{1}{33} + \\frac{1}{34} + \\frac{1}{35} + \\frac{1}{36} + \\frac{1}{37} + \\frac{1}{38} + \\frac{1}{39} + \\frac{1}{40} + \\frac{1}{41} + \\frac{1}{42} + \\frac{1}{43} + \\frac{1}{44} + \\frac{1}{45} + \\frac{"}, {"kind": "sampled", "answer": "1. For $1 \\le n \\le 10$, the numbers from 1 to $10^n$ are in an arithmetic progression with a first term of 1 and a common difference of 1.\n2. The sum of the reciprocals of the non-zero digits of these numbers is $S_n = \\frac{1}{2^n}$.\n3. For numbers with $n$ or more digits, since the reciprocal of the number is less than 1, we must have $S_n < 1$.\n4. Since the reciprocals of the non-zero digits are less than $1$, we need at least $1 \\cdot 10^{n-1} + 10^n$ to adjust the sum.\n\nLet's calculate the sum for $n = 1$:\n$$S_{1} = \\frac{1}{2} + \\frac{1}{2} + \\frac{1}{2} + \\frac{1}{2} +... + \\frac{1}{2}$$\nUsing the formula for the sum of an arithmetic series, we find:\n$$S_{1} = \\frac{1}{2} \\times (1 + 2 + 3 +... + 10)$$\nThe sum of the first 10 natural numbers is:\n$$S_{10} = \\frac{10(1 + 2 + 3 +... + 10)}{2} = 55$$\n\nFor $n = 2$:\n$$S_{2} = \\frac{1}{2^2} + \\frac{1}{2^2} + \\frac{1}{2^2} + \\frac{1}{2^2}$$\nUsing the formula for the sum of an arithmetic series, we find:\n$$S_{2} = \\frac{1}{2^2} \\times (1 + 2 + 3 +... + 20)$$\nThe sum of the first 20 natural numbers is:\n$$S_{20} = \\frac{20(1 + 2 + 3 +... + 20)}{2} = 280$$\n\nFor $n = 3$:\n$$S_{3} = \\frac{1}{2^3} + \\frac{1}{2"}]} +{"index": 85, "row": {"problem": "Example 12. Given $x=19^{94}-1, y=2^{m} \\cdot 3^{n} \\cdot 5^{l}$ $(m, n, l$ are non-negative integers, and $m+n+l \\neq 0)$. Find the sum $S$ of all divisors of $x$ that are of the form $y$.", "solution": "$$\n\\begin{array}{l}\n\\text { Solve } x=(20-1)^{94}-1 \\\\\n=20^{94}-C_{94}^{1} \\cdot 20^{93}+\\cdots-C_{94}^{93} \\cdot 20 \\\\\n=2^{3}\\left(2 n_{1}-235\\right) \\\\\n=2^{3}\\left[2\\left(n_{1}-118\\right)+1\\right] .\\left(n_{1} \\in \\mathbb{N}\\right) \\\\\nx=(18+1)^{94}-1 \\\\\n=18^{94}+C_{94}^{1} \\cdot 18^{93}+\\cdots+C_{94}^{93} \\cdot 18 \\\\\n=3^{2}\\left(3 n_{2}+188\\right) \\\\\n=3^{2}\\left[3\\left(n_{2}+62\\right)+2\\right] .\\left\\{n_{2} \\in \\mathbb{N}\\right) \\\\\nx=(20-1)^{94}-1 \\\\\n=20^{94}-C_{94}^{1} \\cdot 20^{93}+\\cdots+C_{94}^{92} \\cdot 20^{2} \\\\\n-C_{94}^{93} \\cdot 20 \\\\\n=5\\left(5 n_{3}+376\\right) \\\\\n=5\\left[5\\left(n_{3}+75\\right)+1\\right] .\\left(n_{3} \\in \\mathbb{N}\\right) \\\\\n\\end{array}\n$$\n$$\n\\therefore y=2^{m} \\cdot 3^{n} \\cdot 5^{l} \\text { satisfies } 0 \\leqslant m \\leqslant 3,0 \\leqslant n \\leqslant\n$$\n$2,0 \\leqslant l \\leqslant 1$, and $m+n+l \\neq 0$. Therefore,\n$$\n\\begin{aligned}\nS= & \\left(2^{0}+2^{1}+2^{2}+2^{3}\\right)\\left(3^{0}+3^{1}+3^{2}\\right) \\\\\n& \\cdot\\left(5^{0}+5^{1}\\right)-1 \\\\\n= & 1169\n\\end{aligned}\n$$", "answer": "1169", "source": "cn_contest", "problem_type": "Number Theory", "uuid": "0b90b51f-930f-59ea-acdb-d5bf9f3d5d58", "difficulty_band": "medium", "correctness_count": 3}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\nExample 12. Given $x=19^{94}-1, y=2^{m} \\cdot 3^{n} \\cdot 5^{l}$ $(m, n, l$ are non-negative integers, and $m+n+l \\neq 0)$. Find the sum $S$ of all divisors of $x$ that are of the form $y$.", "candidates": [{"kind": "greedy", "answer": "To find the sum of all divisors of $x$ that are of the form $y$, we first need to find the prime factorization of $x$.\n\nGiven $x = 19^{94} - 1$, we can express $19^{94}$ in terms of $19$ and $19^{94}$ itself.\n\nSince $19^{94} = (19^{2})^{94} = (19^{2})^{94} = (19^{2})^{94} = (19^{2})^{94} = (19^{2})^{94} = (19^{2})^{94} = (19^{2})^{94} = (19^{2})^{94} = (19^{2})^{94} = (19^{2})^{94} = (19^{2})^{94} = (19^{2})^{94} = (19^{2})^{94} = (19^{2})^{94} = (19^{2})^{94} = (19^{2})^{94} = (19^{2})^{94} = (19^{2})^{94} = (19^{2})^{94} = (19^{2})^{94} = (19^{2})^{94} = (19^{2})^{94} = (19^{2})^{94} = (19^{2})^{94} = (19^{2})^{94} = (19^{2})^{94} = (19^{2})^{94} = (19^{2})^{94} = (19^{2})^{94} = (19^{2})^{94} = (19^{2})^{94} = (19^{2})^{94} = (19^{2})^{94} = (19^{2})^{94} = (19^{2})^{94} = (19^{2})^{94} = (19^{2})^{94} = (19^{2})^{94} = (19^{2})"}, {"kind": "sampled", "answer": "The number $x=19^{94}-1$ can be rewritten as $2^{19^{94}-1} = 2^{2^{94}} = (2^4)^{2^{94}} = 16^{2^{94}}$.\n\nNow, we can see that $S$ is the sum of all divisors of $x$ that are of the form $y$. To find these divisors, we need to factorize $x$ and the number of divisors.\n\nWe can rewrite $x$ as $2^{2^94} - 1 = (2^4)^{2^94} - 1 = (2^4)^{2^9(64-1)} - 1 = (2^4)^{2^9(63)} - 1$.\n\nThe number of divisors of $x$ that are of the form $y$ is $(63+1) = 64$.\n\nSo, the answer is:\n\\[ \\boxed{64} \\]\n\nFinal answer:\n#### 64"}]} +{"index": 86, "row": {"problem": "8. In a circle, three chords $A A_{1}, B B_{1}, C C_{1}$ intersect at one point. The angular measures of the arcs $A C_{1}, A B, C A_{1}$, and $A_{1} B_{1}$ are $150^{\\circ}, 30^{\\circ}, 60^{\\circ}$, and $30^{\\circ}$, respectively. Find the angular measure of the arc $B_{1} C_{1}$.", "solution": "Solution: Let's formulate several auxiliary statements.\n\n1) Let the angular measure of the arc $AB$ (Fig.1) be $\\varphi$. (This means that $\\varphi$ is equal to the corresponding central angle $AOB$.) Then the length of the chord\n\n![](https://cdn.mathpix.com/cropped/2024_05_06_09f4d2a1d6569731a9d0g-2.jpg?height=311&width=323&top_left_y=907&top_left_x=1569)\n$AB=2R\\sin(\\varphi/2)$. Here $R$ is the radius of the circle.\n\n2) Let two chords $AA_1$ and $BB_1$ intersect at point $T$ (Fig.2). The angular measures of the arcs $AB$ and $A_1B_1$ are $\\varphi$ and $v$. Triangles $ATB$ and $A_1TB_1$ are similar by two angles (equal angles are marked). The similarity ratio\n\n$$\nk=AB / A_1B_1=\\sin(\\varphi/2) / \\sin(v/2)\n$$\n\n![](https://cdn.mathpix.com/cropped/2024_05_06_09f4d2a1d6569731a9d0g-2.jpg?height=254&width=220&top_left_y=1643&top_left_x=193)\n\nFig. 1\n\n![](https://cdn.mathpix.com/cropped/2024_05_06_09f4d2a1d6569731a9d0g-2.jpg?height=411&width=411&top_left_y=1551&top_left_x=480)\n\nFig. 2\n\n![](https://cdn.mathpix.com/cropped/2024_05_06_09f4d2a1d6569731a9d0g-2.jpg?height=480&width=488&top_left_y=1505&top_left_x=955)\n\nFig. 3\n\n3) Referring to Fig. 3. Three chords intersect at one point, denoted as $T$. The angular measures of the six resulting arcs are marked on the figure. From the similarity of triangles $ATB$ and $A_1TB_1$ it follows (see point 2) the equality $AT / B_1T = \\sin(\\varphi/2) / \\sin(v/2)$. Similarly, $\\triangle B T C \\sim \\Delta B_1 T C_1 \\Rightarrow B_1 T / C T = \\sin(\\psi/2) / \\sin(u/2), \\Delta C T A_1 \\sim \\Delta A T C_1 \\Rightarrow C T / A T = \\sin(\\theta/2) / \\sin(w/2)$. Multiplying the last three equalities, we get:\n\n$1 = AT / B_1T \\cdot B_1T / CT \\cdot CT / AT = \\sin(\\varphi/2) / \\sin(v/2) \\cdot \\sin(\\psi/2) / \\sin(u/2) \\cdot \\sin(\\theta/2) / \\sin(w/2)$.\n\nThus, the necessary (and in fact sufficient) condition for three chords to intersect at one point is the equality:\n\n$$\n\\sin(\\varphi/2) \\sin(\\theta/2) \\sin(\\psi/2) = \\sin(u/2) \\sin(v/2) \\sin(w/2)\n$$\n\nNow it is not difficult to obtain the answer to the problem. Substituting the given data into this relation $w=150^\\circ, \\varphi=30^\\circ, \\theta=60^\\circ, v=30^\\circ$, and expressing $u$ from the equality $\\varphi + u + \\theta + v + \\psi + w = 360^\\circ$, we get the equation for determining the desired angle $\\psi$:\n\n$$\n\\sin 15^\\circ \\sin(\\psi/2) \\sin 30^\\circ = \\sin 15^\\circ \\sin \\left(\\left(90^\\circ - \\psi\\right)/2\\right) \\sin 75^\\circ\n$$\n\nFrom this, it is not difficult to obtain that $\\psi = 60^\\circ$.\n\nAnswer: $60^\\circ$.", "answer": "60", "source": "olympiads", "problem_type": "Geometry", "uuid": "8623684b-b730-50d6-9c4a-96567d9d208d", "difficulty_band": "hard", "correctness_count": 1}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\n8. In a circle, three chords $A A_{1}, B B_{1}, C C_{1}$ intersect at one point. The angular measures of the arcs $A C_{1}, A B, C A_{1}$, and $A_{1} B_{1}$ are $150^{\\circ}, 30^{\\circ}, 60^{\\circ}$, and $30^{\\circ}$, respectively. Find the angular measure of the arc $B_{1} C_{1}$.", "candidates": [{"kind": "greedy", "answer": "The problem states that three chords $A A_{1}, B B_{1}, C C_{1}$ intersect at one point. The angular measures of the arcs $A C_{1}, A B, C A_{1}$, and $A_{1} B_{1}$ are $150^{\\circ}, 30^{\\circ}, 60^{\\circ}$, and $30^{\\circ}$, respectively. We need to find the angular measure of the arc $B_{1} C_{1}$.\n\nFirst, let's denote the measure of the arc $B_{1} C_{1}$ as $x$. Since the chords $A A_{1}$, $B B_{1}$, and $C C_{1}$ intersect at one point, the sum of the measures of the arcs $A C_{1}$, $A B$, and $C A_{1}$ is equal to the sum of the measures of the arcs $A B_{1}$, $B C_{1}$, and $A_{1} B_{1}$.\n\nThe sum of the measures of the arcs $A B_{1}$, $B C_{1}$, and $A_{1} B_{1}$ is equal to the sum of the measures of the arcs $A C_{1}$, $A B$, and $C A_{1}$.\n\nTherefore, we have:\n\n$$x = 150^{\\circ} + 30^{\\circ} + 60^{\\circ}$$\n\nSolving for $x$, we get:\n\n$$x = 150^{\\circ} + 30^{\\circ} + 60^{\\circ} = 240^{\\circ}$$\n\nSo, the angular measure of the arc $B_{1} C_{1}$ is $240^{\\circ}$.\n\nFinal answer:\n#### 240"}, {"kind": "sampled", "answer": "This problem primarily tests the application of the angle sum and arc measures. The key to solving this problem is to find the sum of the angles of a triangle, which is a basic question.\n\nFirst, let's find the sum of the angles of triangle $ABC$. This can be done by using the fact that the sum of the angles in a triangle is $180^{\\circ}$.\n\n$$A_{1}B_{1} + A_{1}C_{1} = 180^{\\circ}$$\n\nNext, we can find the sum of the angles of triangle $ABC$. This can be done by using the fact that the sum of the angles in a triangle is $180^{\\circ}$.\n\n$$B_{1}C_{1} + A_{1}B_{1} = 180^{\\circ}$$\n\nNow, we can find the measure of angle $A_{1}B_{1}$. This can be done by using the fact that angle $A_{1}B_{1}$ is an inscribed angle of the circle. Therefore, it measures $30^{\\circ}$.\n\n$$A_{1}B_{1} = 30^{\\circ}$$\n\nNext, we can find the measure of angle $B_{1}C_{1}$. This can be done by using the fact that angle $B_{1}C_{1}$ is an inscribed angle of the circle. Therefore, it measures $60^{\\circ}$.\n\n$$B_{1}C_{1} = 60^{\\circ}$$\n\nFinally, we can find the measure of the arc $B_{1}C_{1}$. This can be done by using the fact that the sum of the measures of angles in a circle is $360^{\\circ}$. Therefore, the measure of the arc $B_{1}C_{1}$ is $360^{\\circ} - 150^{\\circ} - 60^{\\circ} = \\boxed{250^{\\circ}}$.\n\nFinal answer:\n#### 250"}]} +{"index": 87, "row": {"problem": "12.1. Find the smallest natural number consisting of identical digits and divisible by 18.\n\n$$\n(5-6 \\text { cl.) }\n$$", "solution": "12.1. It is clear that the desired number must be even and have a sum of digits divisible by 9. Among two-digit numbers consisting of identical digits, there is no such number. Among three-digit numbers, the number 666 fits this description. It will be the smallest.\n\n保留源文本的换行和格式，翻译结果如下：\n\n12.1. It is clear that the desired number must be even and have a sum of digits divisible by 9. Among two-digit numbers consisting of identical digits, there is no such number. Among three-digit numbers, the number 666 fits this description. It will be the smallest.", "answer": "666", "source": "olympiads", "problem_type": "Number Theory", "uuid": "324bc0e8-7188-5a8b-958e-a8c1b4637648", "difficulty_band": "hard", "correctness_count": 1}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\n12.1. Find the smallest natural number consisting of identical digits and divisible by 18.\n\n$$\n(5-6 \\text { cl.) }\n$$", "candidates": [{"kind": "greedy", "answer": "12.1. The smallest natural number consisting of identical digits and divisible by 18 is 55.\n\nFinal answer:\n#### 55"}, {"kind": "sampled", "answer": "2.1. The smallest natural number consisting of identical digits and divisible by 18 is 69.\n\nFinal answer:\n#### 69"}]} +{"index": 88, "row": {"problem": "1. (Option 1) The decimal representation of the natural number $n$ contains sixty-three digits. Among these digits, there are twos, threes, and fours. No other digits are present. The number of twos is 22 more than the number of fours. Find the remainder when $n$ is divided by 9.", "solution": "# Answer. 5.\n\n(Option 2) The decimal representation of a natural number $n$ contains sixty-one digits. Among these digits, there are threes, fours, and fives. No other digits are present. The number of threes is 11 more than the number of fives. Find the remainder when $n$ is divided by 9.\n\nAnswer. 8.\n\nCriteria. \"干\" The correct answer is obtained by considering a specific case.", "answer": "8", "source": "olympiads", "problem_type": "Number Theory", "uuid": "3852a73f-a1fb-5e95-b0fa-4694789bf6dd", "difficulty_band": "hard", "correctness_count": 1}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\n1. (Option 1) The decimal representation of the natural number $n$ contains sixty-three digits. Among these digits, there are twos, threes, and fours. No other digits are present. The number of twos is 22 more than the number of fours. Find the remainder when $n$ is divided by 9.", "candidates": [{"kind": "greedy", "answer": "The number of twos is 22 more than the number of fours.\n\nThe number of fours is 40 less than the number of twos.\n\nThe number of twos is 22 more than the number of fours, so it is 22 + 40 = 62.\n\nThe number of fours is 40 less than the number of twos, so it is 62 - 40 = 22.\n\nThe number of fours is 22 more than the number of twos, so it is 62 + 22 = 84.\n\nThe number of twos is 62, so the number of fours is 84 - 62 = 22.\n\nThe number of fours is 22 more than the number of twos, so it is 22 + 22 = 44.\n\nThe number of twos is 62, so the number of fours is 44 - 62 = -18.\n\nSince the number of fours is negative, the number of fours is -18.\n\nThe number of twos is 62, so the number of fours is -18 + 62 = 44.\n\nThe number of fours is 44, so the number of twos is 44 - 44 = 0.\n\nThe number of twos is 0, so the number of fours is 0 + 44 = 44.\n\nThe number of fours is 44, so the number of twos is 44 - 44 = 0.\n\nThe number of twos is 0, so the number of fours is 0 + 44 = 44.\n\nThe number of twos is 0, so the number of fours is 0 + 44 = 44.\n\nThe number of twos is 0, so the number of fours is 0 + 44 = 44.\n\nThe number of twos is 0, so the number of fours is 0 + 44 = 44.\n\nThe number of twos is 0, so the number of fours is 0"}, {"kind": "sampled", "answer": "The number of twos in the decimal representation of $n$ can be expressed as $n = 22A + F$, where $A$ is the number of digits of $n$ and $F$ is the number of twos.\n\nWe know that $22A + F = 63$, and we know that there are 44 twos in the decimal representation of $n$.\n\nSince $n = 22A + F$ and $n$ is a multiple of 9, we know that $n \\equiv 0 \\pmod{9}$.\n\nSo, $22A + F \\equiv 0 \\pmod{9}$.\n\nThe remainder when $n$ is divided by 9 can be found by taking the remainder of 63 when divided by 9.\n\nThe remainder of 63 divided by 9 is 0, since $63 = 9 \\cdot 7 + 8$.\n\nTherefore, the remainder when $n$ is divided by 9 is 8.\n\nFinal answer:\n#### 8"}]} +{"index": 89, "row": {"problem": "Example 7 A shipping company has a ship leaving Harvard for New York every noon, and at the same time every day, a ship also leaves New York for Harvard. It takes seven days and seven nights for the ships to complete their journeys in both directions, and they all sail on the same route. How many ships of the same company will the ship leaving Harvard at noon today encounter on its way to New York?", "solution": "Solution 1: Establishing Equal Relationships of Distance and Time.\nAnalysis: If this ship meets another ship opposite to it at point $A$, then the next ship opposite to it is at point $B$. By the equal relationship of distance, it is easy to know that they meet at the midpoint of $A B$, and the interval is exactly half a day. Therefore, throughout the process, it should meet 15 ships coming from the opposite direction.\nSolution 2: Establishing Indeterminate Equation Method.\nLet the ship that departs from Harvard at noon today meet a ship that departed $x$ days ago, then\n\\[\n\\begin{array}{l}\nx+y=7(\\hat{u} \\leqslant x \\leqslant 7), \\\\\nx-y=n(n \\in Z). \\\\\n\\therefore 2 x=7+n. \\\\\n\\text { Thus, }-7 \\leqslant n \\leqslant 7.\n\\end{array}\n\\]\n\nTherefore, $n$ can take $-7,-6, \\cdots,-1,0,1, \\cdots, 6,7$. There are a total of 15 ships.", "answer": "15", "source": "cn_contest", "problem_type": "Logic and Puzzles", "uuid": "c1781d85-af5d-50ae-ba7c-4d13e87b7ea8", "difficulty_band": "hard", "correctness_count": 1}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\nExample 7 A shipping company has a ship leaving Harvard for New York every noon, and at the same time every day, a ship also leaves New York for Harvard. It takes seven days and seven nights for the ships to complete their journeys in both directions, and they all sail on the same route. How many ships of the same company will the ship leaving Harvard at noon today encounter on its way to New York?", "candidates": [{"kind": "greedy", "answer": "The ship leavingin New York will encounter the ship leaving in New York on the same day.\n\nFinal answer: yes"}, {"kind": "sampled", "answer": "The ship leaving Farid will encounter:\n\\[ \\boxed{7} \\]\n\nFinal answer:\n#### 7"}]} +{"index": 90, "row": {"problem": "Point $P$ is on the line $y=5 x+3$. The coordinates of point $Q$ are $(3,-2)$. If $M$ is the midpoint of $P Q$, then $M$ must lie on the line\n(A) $y=\\frac{5}{2} x-\\frac{7}{2}$\n(B) $y=5 x+1$\n(C) $y=-\\frac{1}{5} x-\\frac{7}{5}$\n(D) $y=\\frac{5}{2} x+\\frac{1}{2}$\n(E) $y=5 x-7$\n\n## Part C: Each correct answer is worth 8.", "solution": "Point $P$ is on the line $y=5 x+3$. The coordinates of point $Q$ are $(3,-2)$. If $M$ is the midpoint of $P Q$, then $M$ must lie on the line\n(A) $y=\\frac{5}{2} x-\\frac{7}{2}$\n(B) $y=5 x+1$\n(C) $y=-\\frac{1}{5} x-\\frac{7}{5}$\n(D) $y=\\frac{5}{2} x+\\frac{1}{2}$\n(E) $y=5 x-7$\n\nWe start by drawing a diagram and labelling the intercepts.\n\n## Solution 1\n\nSince the point $P$ is on the line $y=5 x+3$, select $P(0,3)$ as a point on this line.\n\nThe midpoint of $P Q$ is $M\\left(\\frac{3+0}{2}, \\frac{-2+3}{2}\\right)=M\\left(\\frac{3}{2}, \\frac{1}{2}\\right)$.\n\nThe required line must contain $M$ and be midway between the given point and $y=5 x+3$. The only possible line meeting this requirement is the line containing $M\\left(\\frac{3}{2}, \\frac{1}{2}\\right)$ and which has a slope of 5 . The required line will this have as its equation\n\n$$\n\\begin{aligned}\ny-\\frac{1}{2} & =5\\left(x-\\frac{3}{2}\\right) \\\\\n\\text { or, } \\quad y & =5 x-7\n\\end{aligned}\n$$\n\n![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-266.jpg?height=876&width=656&top_left_y=337&top_left_x=1293)\n\n## Solution 2\n\nLet a general point on the line $y=5 x+3$ be represented by $(a, 5 a+3)$. Also, let a point on the required line be $M(x, y)$. Since $M(x, y)$ is the midpoint of $P Q$ then\n\n(1) $x=\\frac{a+3}{2}$ and\n\n(2) $y=\\frac{(5 a+3)+(-2)}{2}$\n\n$$\ny=\\frac{5 a+1}{2}\n$$\n\nSolving (1) for $a$, we have $a=2 x-3$ and solving (2) for $a$, we have $\\frac{2 y-1}{5}=a$.\n\nEquating gives, $2 x-3=\\frac{2 y-1}{5}$\n\n$$\n\\begin{aligned}\n10 x-15 & =2 y-1 \\\\\n\\text { or, } \\quad y & =5 x-7\n\\end{aligned}\n$$\n\nANSWER: (E)\n\n## Part C", "answer": "5x-7", "source": "olympiads", "problem_type": "Algebra", "uuid": "3e0ae3fc-b8b9-5b46-9138-7902cd524593", "difficulty_band": "medium", "correctness_count": 2}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\nPoint $P$ is on the line $y=5 x+3$. The coordinates of point $Q$ are $(3,-2)$. If $M$ is the midpoint of $P Q$, then $M$ must lie on the line\n(A) $y=\\frac{5}{2} x-\\frac{7}{2}$\n(B) $y=5 x+1$\n(C) $y=-\\frac{1}{5} x-\\frac{7}{5}$\n(D) $y=\\frac{5}{2} x+\\frac{1}{2}$\n(E) $y=5 x-7$\n\n## Part C: Each correct answer is worth 8.", "candidates": [{"kind": "greedy", "answer": "The midpoint $M$ of $PQ$ is given by:\n$$M = \\left(\\frac{x_1+x_2}{2}, \\frac{y_1+y_2}{2}\\right)$$\n\nSubstituting the given values for $P$ and $Q$ into the midpoint formula, we get:\n$$M = \\left(\\frac{3+3}{2}, \\frac{-2+(-2)}{2}\\right)$$\n\nSimplifying the coordinates, we get:\n$$M = \\left(\\frac{3}{2}, -2\\right)$$\n\nTherefore, the coordinates of point $M$ are $\\left(\\frac{3}{2}, -2\\right)$.\n\nThe correct answer is:\n(A) $\\boxed{y=\\frac{5}{2} x-\\frac{7}{2}}$\n\nFinal answer:\n(A)"}, {"kind": "sampled", "answer": "To find the $y$-coordinate of point $M$, we can use the midpoint formula. However, we should notice that since $M$ is the midpoint, it lies on the line segment $\\overline{PQ}$. \n\nLet $t$ be the $x$-coordinate of $M$. Since the coordinates of $M$ are the average of the coordinates of $P$ and $Q$, we can express $PQ$ as a function of $t$. The equation of line $PQ$ is given by:\n\n$$y = 5x + 3$$\n\nThe midpoint formula gives us the $x$-coordinate of $M$ as the average of the $x$-coordinates of $P$ and $Q$.\n\nThe coordinates of $M$ are:\n\n$$x_{M} = \\frac{t + (3 - t)}{2}$$\n\nThe midpoint formula gives us the $y$-coordinate of $M$ as the average of the $y$-coordinates of $P$ and $Q$.\n\nTherefore, we have:\n\n$$y_{M} = \\frac{5x_{M} + 3}{2}$$\n\nWe can simplify this expression by combining the terms:\n\n$$y_{M} = \\frac{5}{2}x_{M} + \\frac{3}{2}$$\n\nThis expression represents the $y$-coordinate of $M$.\n\nSo, the correct answer is:\n\n$$y_{M} = \\boxed{y_{M}}$$\n#### 5"}]} +{"index": 91, "row": {"problem": "Find the largest positive real $ k$, such that for any positive reals $ a,b,c,d$, there is always:\r\n\\[ (a\\plus{}b\\plus{}c) \\left[ 3^4(a\\plus{}b\\plus{}c\\plus{}d)^5 \\plus{} 2^4(a\\plus{}b\\plus{}c\\plus{}2d)^5 \\right] \\geq kabcd^3\\]", "solution": "1. We start by considering the given inequality:\n \\[\n (a+b+c) \\left[ 3^4(a+b+c+d)^5 + 2^4(a+b+c+2d)^5 \\right] \\geq kabcd^3\n \\]\n We need to find the largest positive real \\( k \\) such that this inequality holds for any positive reals \\( a, b, c, d \\).\n\n2. Let's test the inequality with specific values. Set \\( a = b = c = \\frac{1}{3} \\) and \\( d = 1 \\). Then:\n \\[\n a + b + c = 1\n \\]\n Substituting these values into the inequality, we get:\n \\[\n 1 \\left[ 3^4(1+1)^5 + 2^4(1+2)^5 \\right] \\geq k \\left( \\frac{1}{3} \\cdot \\frac{1}{3} \\cdot \\frac{1}{3} \\cdot 1^3 \\right)\n \\]\n Simplifying the right-hand side:\n \\[\n k \\left( \\frac{1}{27} \\right) = \\frac{k}{27}\n \\]\n\n3. Now, we simplify the left-hand side:\n \\[\n 3^4 \\cdot 2^5 + 2^4 \\cdot 3^5 = 81 \\cdot 32 + 16 \\cdot 243\n \\]\n Calculating each term:\n \\[\n 81 \\cdot 32 = 2592\n \\]\n \\[\n 16 \\cdot 243 = 3888\n \\]\n Adding these together:\n \\[\n 2592 + 3888 = 6480\n \\]\n\n4. Therefore, the inequality becomes:\n \\[\n 6480 \\geq \\frac{k}{27}\n \\]\n Solving for \\( k \\):\n \\[\n k \\leq 6480 \\cdot 27 = 174960\n \\]\n\n5. To prove that this value of \\( k \\) is indeed the maximum, we use the AM-GM inequality. Assume \\( a + b + c = 1 \\). By the AM-GM inequality:\n \\[\n 3^4(1+d)^5 + 2^4(1+2d)^5 \\geq 3^4(2\\sqrt{d})^5 + 2^4(3\\sqrt[3]{d^2})^5\n \\]\n Simplifying the right-hand side:\n \\[\n 3^4 \\cdot 32d^{5/2} + 2^4 \\cdot 243d^{10/3}\n \\]\n Using the weighted AM-GM inequality:\n \\[\n 6^4 \\cdot 5 \\left( \\frac{2}{5}d^{5/2} + \\frac{3}{5}d^{10/3} \\right) \\geq 6^4 \\cdot 5 \\cdot d^3\n \\]\n Since \\( a + b + c = 1 \\), we have:\n \\[\n 6^4 \\cdot 5 \\cdot d^3 \\geq 6^4 \\cdot 5 \\cdot \\frac{abc}{27} d^3\n \\]\n Therefore:\n \\[\n k = \\frac{6^4 \\cdot 5}{27} = 174960\n \\]\n\nThe final answer is \\( \\boxed{ k = 174960 } \\)", "answer": "k = 174960", "source": "aops_forum", "problem_type": "Inequalities", "uuid": "0acb9d95-6a7b-53e6-b8d7-b2f83f1b143d", "difficulty_band": "medium", "correctness_count": 2}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\nFind the largest positive real $ k$, such that for any positive reals $ a,b,c,d$, there is always:\r\n\\[ (a\\plus{}b\\plus{}c) \\left[ 3^4(a\\plus{}b\\plus{}c\\plus{}d)^5 \\plus{} 2^4(a\\plus{}b\\plus{}c\\plus{}2d)^5 \\right] \\geq kabcd^3\\]", "candidates": [{"kind": "greedy", "answer": "To find the largest positive real $k$, we need to analyze the given inequality and understand the conditions under which it holds true.\n\nThe inequality involves a product of two terms, each involving a sum of cubes and a fifth power. The left-hand side (LHS) is a sum of terms, each of which involves a sum of fourth powers and a fifth power. The right-hand side (RHS) is a constant $kabcd^3$.\n\nTo simplify the problem, we can use the AM-GM (Arithmetic Mean-Geometric Mean) inequality, which states that for non-negative real numbers $x_1, x_2, \\ldots, x_n$, the following inequality holds:\n\\[ \\frac{x_1 + x_2 + \\cdots + x_n}{n} \\geq \\sqrt[n]{x_1x_2\\cdots x_n} \\]\n\nApplying this to the terms in the LHS, we get:\n\\[ \\frac{a + b + c + d}{4} \\geq \\sqrt[4]{a^4b^4c^4d^4} \\]\n\nThis implies that:\n\\[ a + b + c + d \\geq 4\\sqrt[4]{a^4b^4c^4d^4} \\]\n\nNow, let's consider the RHS of the inequality. We can rewrite the RHS as:\n\\[ kabcd^3 = k(a + b + c)(b + c)(c + d)(d + 3) \\]\n\nTo find the largest possible value of $k$, we need to find the maximum value of the RHS.\n\nUsing the AM-GM inequality again, we have:\n\\[ a + b + c + d \\geq 4\\sqrt[4]{a^4b^4c^4d^4} \\geq 4\\sqrt[4]{(abcd)^4} = 4\\sqrt[4]{(abcd)^4} = 4\\sqrt[4]{(abcd)^4} = 4\\sqrt[4]{(abcd)^4} = 4\\sqrt[4]{(abcd)^4} = 4\\sqrt[4]{(abcd)^4} = 4\\sqrt[4]{"}, {"kind": "sampled", "answer": "To find the largest positive real $k$, we will use mathematical induction.\n\n#### Base Case:\nFor $n = 1$, the statement becomes:\n\\[ (a+b) \\left[ 3^4(a+b+c) \\right] = 3^4a + 3^4b + 3^4c + 2^4a + 2^4b + 2^4c + 2^4d \\geq kabcd^3 \\]\n\nSubstituting $a = b = c = d = 1$:\n\\[ 3^4 = 81 \\]\n\\[ 81 + 3^4 + 2^4 = 81 + 81 + 16 = 168 \\geq 168k \\]\nThis holds true.\n\n#### Inductive Step:\nAssume that the statement holds for $n = k$. Consider $n = k+1$:\n\\[ [(a+b) \\left[ 3^4(a+b+c) \\right] + 2^4(a+b+c+2d)^5] \\geq (k+1)abcd^3 + 2^4(a+b+c+2d)^5 \\]\n\nUsing the inductive hypothesis and the inequality from the base case, we get:\n\\[ (a+b) \\left[ 3^4(a+b+c) + 2^4(a+b+c+2d)^5 \\right] \\geq (k+1)abcd^3 + 2^4(a+b+c+2d)^5 \\]\n\nSubtracting $2^4(a+b+c+2d)^5$ from both sides:\n\\[ (a+b) \\left[ 3^4(a+b+c) + 2^4(a+b+c+2d)^5 - 2^4(a+b+c+2d)^5 = (k+1)abcd^3 \\right] \\geq (k+1)abcd^3 \\]\n\nThis simplifies to:\n\\[ (a+b) \\left[ 3^4(a+b+c)"}]} +{"index": 92, "row": {"problem": "Example 6 Find the unit digit of the sum $1^{2}+2^{2}+3^{2}+4^{2}+\\cdots+1994^{2}$.", "solution": "Solution: Since this problem only requires the unit digit of the sum, we only need to consider the unit digit of each number. Thus, the original problem simplifies to finding the unit digit of\n$$\n\\underbrace{i^{2}+2^{2}+3^{2}+4^{2}+\\cdots+9^{2}}_{\\text {199 groups }}+1^{2}+2^{2}+3^{2}+4^{2}\n$$\n\nThe unit digits follow a periodic pattern:\n$$\n1,4,9,6,5,6,9,4,1 \\text {, }\n$$\n\nThe unit digit of the sum of each group is 5, so the unit digit of the sum of 199 groups is also 5. Considering the unit digit sum of the remaining four numbers is 0, the unit digit of the required sum is 5.", "answer": "5", "source": "cn_contest", "problem_type": "Number Theory", "uuid": "0277fd2d-31ad-5b45-9337-9343a842c002", "difficulty_band": "medium", "correctness_count": 2}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\nExample 6 Find the unit digit of the sum $1^{2}+2^{2}+3^{2}+4^{2}+\\cdots+1994^{2}$.", "candidates": [{"kind": "greedy", "answer": "The unit digit of the sum $1^{2}+2^{2}+3^{2}+4^{2}+\\cdots+1994^{2}$ is 1.\n\nFinal answer:\n#### 1"}, {"kind": "sampled", "answer": "To find the unit digit of the sum $1^{2}+2^{2}+3^{2}+4^{2}+\\cdots+1994^{2}$, we first need to understand the pattern of the unit digits of the squares of numbers from $1$ to $1994$.\n\nNotice that the unit digit of $1^2$ is $1$. The unit digit of $2^2$ is $4$, the unit digit of $3^2$ is $9$, and the unit digit of $4^2$ is $6$.\n\nThe pattern in the unit digits repeats every $4$ numbers, starting from $4^2$. Therefore, we can find the unit digit of the sum $1^{2}+2^{2}+3^{2}+4^{2}+\\cdots+1994^{2}$ by finding the remainder when $1994$ is divided by $4$, and then adding $1$.\n\nThe remainder of $1994$ divided by $4$ is $3$ (since $1994 = 4 \\times 498 + 2$). Therefore, the unit digit of the sum is $1 + 2 = 3$.\n\nSo, the unit digit of the sum $1^{2}+2^{2}+3^{2}+4^{2}+\\cdots+1994^{2}$ is $3$.\nFinal answer:\n#### \\boxed{3}"}]} +{"index": 93, "row": {"problem": "9. As shown in Figure 2, in quadrilateral $ABCD$, $AB=BC=CD$, $\\angle ABC=78^{\\circ}$, $\\angle BCD=162^{\\circ}$. Let the intersection point of line $AD$ and $BC$ be $E$. Then the size of $\\angle AEB$ is", "solution": "$9.21^{\\circ}$.\nAs shown in Figure 5, draw $D O / / C B$, and\nmake $D O=C B$, connect\n$O A$ and $O B$.\nSince $B C=C D$,\nthus, quadrilateral $B C D O$ is a rhombus. Therefore,\n$$\n\\begin{array}{l}\nA B=B C=B O . \\\\\n\\text { Since } \\angle A B O=78^{\\circ}-\\left(180^{\\circ}-162^{\\circ}\\right)=60^{\\circ},\n\\end{array}\n$$\n\nTherefore, $\\triangle A B O$ is an equilateral triangle.\nLet $\\angle A E B=\\alpha$. Then $\\angle 1=\\angle 2=\\alpha$.\nBy the sum of the interior angles of a quadrilateral being $360^{\\circ}$, we get\n$$\n\\begin{array}{l}\n78^{\\circ}+162^{\\circ}+\\left(60^{\\circ}+\\alpha\\right)+\\left(180^{\\circ}-162^{\\circ}+\\alpha\\right) \\\\\n=360^{\\circ} .\n\\end{array}\n$$\n\nSolving for $\\alpha$ gives $\\alpha=21^{\\circ}$.", "answer": "21^{\\circ}", "source": "cn_contest", "problem_type": "Geometry", "uuid": "5ce24965-2da7-51bb-9592-e3a730bc1de6", "difficulty_band": "hard", "correctness_count": 1}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\n9. As shown in Figure 2, in quadrilateral $ABCD$, $AB=BC=CD$, $\\angle ABC=78^{\\circ}$, $\\angle BCD=162^{\\circ}$. Let the intersection point of line $AD$ and $BC$ be $E$. Then the size of $\\angle AEB$ is", "candidates": [{"kind": "greedy", "answer": "10. In quadrilateral $ABCD$, $AB=BC=CD$, $\\angle ABC=78^{\\circ}$, $\\angle BCD=162^{\\circ}$. Let the intersection point of line $AD$ and $BC$ be $E$. Then the size of $\\angle AEB$ is $162^{\\circ}$.\n\nFinal answer:\n#### 162"}, {"kind": "sampled", "answer": "The solution is to find the angle between $AB$ and $CD$, and the angle between $BC$ and $CD$.\n\nIn $\\triangle ABC$, we have $AB=BC=CD$, hence $\\triangle ACD$ is isosceles, so $\\angle ACD=180^{\\circ}-\\angle ADC=180^{\\circ}-78^{\\circ}=102^{\\circ}$. Since $\\angle BCD=162^{\\circ}$, we have $\\angle CBD=162^{\\circ}-102^{\\circ}=60^{\\circ}$. Since $\\angle ADB=102^{\\circ}$, we have $\\angle ADB=\\angle CBD\\cdot \\frac{180^{\\circ}-\\angle CBD}{2}=\\boxed{36^{\\circ}}$."}]} +{"index": 94, "row": {"problem": "A2. We throw two dice at once multiple times and each time we add the numbers of dots on the top faces of both. What is the maximum number of different sums of dots we can get?\n(A) 10\n(B) 11\n(C) 12\n(D) 18\n(E) 20", "solution": "A2. Possible sums of the dots on the upper faces of the dice are: $2=1+1,3=1+2=2+1$, $4=1+3=2+2=3+1,5=1+4=2+3=3+2=4+1,6=1+5=$ $2+4=3+3=4+2=5+1,7=1+6=2+5=3+4=4+3=5+2=6+1$, $8=2+6=3+5=4+4=5+3=6+2,9=3+6=4+5=5+4=6+3$, $10=4+6=5+5=6+4,11=5+6=6+5$ and $12=6+6$, a total of 11 different sums.", "answer": "11", "source": "olympiads", "problem_type": "Combinatorics", "uuid": "70ad22d4-945b-5910-9b39-8508e55a8b03", "difficulty_band": "medium", "correctness_count": 2}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\nA2. We throw two dice at once multiple times and each time we add the numbers of dots on the top faces of both. What is the maximum number of different sums of dots we can get?\n(A) 10\n(B) 11\n(C) 12\n(D) 18\n(E) 20", "candidates": [{"kind": "greedy", "answer": "When you throw two dice, the maximum number of different sums you can get is 11.\n\nFinal answer: (B) 11"}, {"kind": "sampled", "answer": "A6. The sum of the numbers on a single die is 1+2+3+4+5+6+7+8+9+10+11+12+13+14+15 = 99.\n\nThe maximum number of different sums of dots is 99.\n\nFinal answer: (E)"}]} +{"index": 95, "row": {"problem": "Natural numbers $k, l,p$ and $q$ are such that if $a$ and $b$ are roots of $x^2 - kx + l = 0$ then $a +\\frac1b$ and $b + \\frac1a$ are the roots of $x^2 -px + q = 0$. What is the sum of all possible values of $q$?", "solution": "1. Assume the roots of the equation \\(x^2 - kx + l = 0\\) are \\(\\alpha\\) and \\(\\beta\\). By Vieta's formulas, we have:\n \\[\n \\alpha + \\beta = k \\quad \\text{and} \\quad \\alpha \\beta = l\n \\]\n\n2. We need to find an equation whose roots are \\(\\alpha + \\frac{1}{\\beta}\\) and \\(\\beta + \\frac{1}{\\alpha}\\). The sum of these roots is:\n \\[\n \\left(\\alpha + \\frac{1}{\\beta}\\right) + \\left(\\beta + \\frac{1}{\\alpha}\\right) = \\alpha + \\beta + \\frac{1}{\\alpha} + \\frac{1}{\\beta}\n \\]\n Using the relationships from Vieta's formulas, we can rewrite this as:\n \\[\n \\alpha + \\beta + \\frac{1}{\\alpha} + \\frac{1}{\\beta} = k + \\frac{1}{\\alpha} + \\frac{1}{\\beta}\n \\]\n Since \\(\\frac{1}{\\alpha} + \\frac{1}{\\beta} = \\frac{\\alpha + \\beta}{\\alpha \\beta}\\), we get:\n \\[\n k + \\frac{\\alpha + \\beta}{\\alpha \\beta} = k + \\frac{k}{l}\n \\]\n Therefore, the sum of the roots is:\n \\[\n k + \\frac{k}{l} = \\frac{k(l + 1)}{l}\n \\]\n\n3. The product of the roots \\(\\alpha + \\frac{1}{\\beta}\\) and \\(\\beta + \\frac{1}{\\alpha}\\) is:\n \\[\n \\left(\\alpha + \\frac{1}{\\beta}\\right)\\left(\\beta + \\frac{1}{\\alpha}\\right) = \\alpha \\beta + \\alpha \\cdot \\frac{1}{\\alpha} + \\beta \\cdot \\frac{1}{\\beta} + \\frac{1}{\\alpha \\beta}\n \\]\n Simplifying, we get:\n \\[\n \\alpha \\beta + 1 + 1 + \\frac{1}{\\alpha \\beta} = l + 2 + \\frac{1}{l}\n \\]\n Therefore, the product of the roots is:\n \\[\n l + 2 + \\frac{1}{l} = \\frac{(l + 1)^2}{l}\n \\]\n\n4. We now have the quadratic equation with roots \\(\\alpha + \\frac{1}{\\beta}\\) and \\(\\beta + \\frac{1}{\\alpha}\\) as:\n \\[\n x^2 - \\left(\\frac{k(l + 1)}{l}\\right)x + \\frac{(l + 1)^2}{l} = 0\n \\]\n\n5. Comparing this with the given equation \\(x^2 - px + q = 0\\), we get:\n \\[\n p = \\frac{k(l + 1)}{l} \\quad \\text{and} \\quad q = \\frac{(l + 1)^2}{l}\n \\]\n\n6. We need to find the sum of all possible values of \\(q\\). Since \\(k, l, p,\\) and \\(q\\) are natural numbers, we need \\(l\\) to be a natural number that makes \\(q\\) a natural number. Let's check the possible values of \\(l\\):\n\n - For \\(l = 1\\):\n \\[\n q = \\frac{(1 + 1)^2}{1} = \\frac{4}{1} = 4\n \\]\n\n - For \\(l = 2\\):\n \\[\n q = \\frac{(2 + 1)^2}{2} = \\frac{9}{2} = 4.5 \\quad \\text{(not a natural number)}\n \\]\n\n - For \\(l = 3\\):\n \\[\n q = \\frac{(3 + 1)^2}{3} = \\frac{16}{3} \\quad \\text{(not a natural number)}\n \\]\n\n - For \\(l = 4\\):\n \\[\n q = \\frac{(4 + 1)^2}{4} = \\frac{25}{4} \\quad \\text{(not a natural number)}\n \\]\n\n - For \\(l = 5\\):\n \\[\n q = \\frac{(5 + 1)^2}{5} = \\frac{36}{5} \\quad \\text{(not a natural number)}\n \\]\n\n From this, we see that the only natural number value for \\(q\\) is when \\(l = 1\\), giving \\(q = 4\\).\n\nThe final answer is \\(\\boxed{4}\\)", "answer": "4", "source": "aops_forum", "problem_type": "Algebra", "uuid": "a963f2e1-855f-52c2-a251-115a3e45ed27", "difficulty_band": "hard", "correctness_count": 1}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\nNatural numbers $k, l,p$ and $q$ are such that if $a$ and $b$ are roots of $x^2 - kx + l = 0$ then $a +\\frac1b$ and $b + \\frac1a$ are the roots of $x^2 -px + q = 0$. What is the sum of all possible values of $q$?", "candidates": [{"kind": "greedy", "answer": "Since $a$ and $b$ are roots of $x^2 - kx + l = 0$, we have $a + b = k$ and $b + a = l$.\n\nSince $a + b = k$ and $b + a = l$, we have $a = k - b$ and $a = l - b$.\n\nSince $a = k - b$ and $a = l - b$, we have $k - b = l - b$, which implies $k = l$.\n\nSince $a = k - b$ and $a = l - b$, we have $k - b = l - b$, which implies $k = l$.\n\nSince $a = k - b$ and $a = l - b$, we have $k - b = l - b$, which implies $k = l$.\n\nSince $a = k - b$ and $a = l - b$, we have $k - b = l - b$, which implies $k = l$.\n\nSince $a = k - b$ and $a = l - b$, we have $k - b = l - b$, which implies $k = l$.\n\nSince $a = k - b$ and $a = l - b$, we have $k - b = l - b$, which implies $k = l$.\n\nSince $a = k - b$ and $a = l - b$, we have $k - b = l - b$, which implies $k = l$.\n\nSince $a = k - b$ and $a = l - b$, we have $k - b = l - b$, which implies $k = l$.\n\nSince $a = k - b$ and $a = l - b$, we have $k - b = l - b$, which implies $k = l$.\n\nSince $a = k - b$ and $a = l - b$, we have $k - b = l - b$, which implies $k = l$.\n\nSince $a = k - b$ and $a = l - b$, we have $k - b = l - b$, which implies $k = l$.\n\nSince $a = k - b$ and $a = l - b$, we have $k - b = l - b$, which implies $k = l$.\n\nSince $"}, {"kind": "sampled", "answer": "Given that $a$ and $b$ are the roots of $x^2 - kx + l = 0$, we know that:\n\\[ a + \\frac{1}{b} = k \\quad \\text{and} \\quad b + \\frac{1}{a} = l \\]\n\nWe are also given that $a + \\frac{1}{b} = p$ and $b + \\frac{1}{a} = q$. These can be rewritten as:\n\\[ a + \\frac{1}{b} = p \\Rightarrow a = p - \\frac{1}{q} \\]\n\\[ b + \\frac{1}{a} = q \\Rightarrow b = q - \\frac{1}{p} \\]\n\nNow, let's find $q$:\n\\[ a + \\frac{1}{b} = p \\Rightarrow p - \\frac{1}{q} + \\frac{1}{b} = p \\Rightarrow \\frac{1}{b} - \\frac{1}{p} = 0 \\Rightarrow \\frac{1}{p} = \\frac{1}{b} \\Rightarrow b = p \\]\n\n\\[ b + \\frac{1}{a} = q \\Rightarrow q - \\frac{1}{q} + \\frac{1}{a} = q \\Rightarrow \\frac{1}{a} - \\frac{1}{q} = 0 \\Rightarrow \\frac{1}{a} = \\frac{1}{q} \\Rightarrow q = a \\]\n\nSince $b = p$ and $a = p - \\frac{1}{q}$, we have:\n\\[ p - \\frac{1}{p} = q \\Rightarrow p^2 - qp + 1 = 0 \\]\n\nSolve for $p$:\n\\[ p^2 - pq + 1 = 0 \\Rightarrow (p - q)(p + q) + 1 = 0 \\Rightarrow (p - q)(p + q + 1) = 0 \\]\n\n\\[ p = q + 1 \\]\n\nNow, we can find $q$:\n\\[ q = a = p + 1 = (q + 1) + 1 = q + 2 \\"}]} +{"index": 96, "row": {"problem": "13.266. Two athletes start running simultaneously - the first from $A$ to $B$, the second from $B$ to $A$. They run at different but constant speeds and meet at a distance of 300 m from $A$. After running the entire path $A B$, each immediately turns around and meets the other at a distance of 400 m from $B$. Find the length of $A B$.", "solution": "Solution.\n\nConsider the graphs of the runs of two athletes (Fig. 13.9). Let $AB = x$ m, $C$ and $D$ be the points of the first and second meetings, $v_{1}$ and $v_{2}$ be the speeds of the first and second athlete. Then $\\frac{300}{v_{1}}=\\frac{x-300}{v_{2}}\\left({ }^{*}\\right)$. During the time between the meetings $(KL)$, the athletes ran:\n\nthe first $(x-300)+400=x+100$ m;\n\nthe second $300+(x-400)=x-100$ m.\n\nThus, $\\frac{x+100}{v_{1}}=\\frac{x-100}{v_{2}}(* *)$.\n\nFrom $\\left({ }^{*}\\right)$ and (**) we get $\\frac{x+100}{300}=\\frac{x-100}{x-300} \\Rightarrow x=500 \\mathrm{~m}$.\n\nAnswer: 500 m.", "answer": "500\\mathrm{~}", "source": "olympiads", "problem_type": "Algebra", "uuid": "4420a38b-4b83-5df9-9c3b-c67f267d784c", "difficulty_band": "hard", "correctness_count": 1}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\n13.266. Two athletes start running simultaneously - the first from $A$ to $B$, the second from $B$ to $A$. They run at different but constant speeds and meet at a distance of 300 m from $A$. After running the entire path $A B$, each immediately turns around and meets the other at a distance of 400 m from $B$. Find the length of $A B$.", "candidates": [{"kind": "greedy", "answer": "13.266. The first athlete runs from $A$ to $B$, and the second athlete runs from $B$ to $A$. They meet at a distance of 300 m from $A$. After running the entire path $A B$, each immediately turns around and meets the other at a distance of 400 m from $B$. Find the length of $AB$.\n\nFinal answer:\n#### 400"}, {"kind": "sampled", "answer": "13.266. The two athletes start running at the same time, running towards each other, at a distance of 300 m from A. The first athlete runs at a constant speed, and the second athlete runs in the same direction, at a constant speed. The second athlete starts the journey at a distance of 400 m from B.\n\nFrom the problem, we know that the second athlete's speed is 300 m/s, which means the distance the second athlete has to cover is 2000 m. Therefore, the second athlete's speed is 300 m/s.\n\nAs the first athlete runs towards the second athlete, their speeds are added together, so their speeds are 300 + 2000 = 2300 m/s. Therefore, the second athlete runs for 400 seconds, and the distance covered by the second athlete is 10000 m.\n\nThus, the length of the second athlete's path is 10000 / 2300 = 40 m.\n\nTherefore, the total length of the path that the two athletes have to run is 2300 + 40 = 2340 m."}]} +{"index": 97, "row": {"problem": "8. (10 points) 12 Smurfs are sitting around a round table, each Smurf hates the 2 Smurfs sitting next to him, but does not hate the other 9 Smurfs. Papa Smurf needs to send out a team of 5 Smurfs to rescue Smurfette, who was captured by Gargamel, and the team cannot include Smurfs who hate each other. There are $\\qquad$ ways to form the team.", "solution": "【Solution】Solution: According to the requirements, divide into three major categories:\nOne category is to select all odd-numbered ones, the number of groups is $C_{6}^{5}=6$,\nThe second category is to select all even-numbered ones, the number of groups is $C_{6}^{5}=6$,\nThe third category is a mix of odd and even numbers, which is more complex and further divided into 4 subcategories:\n1st category: 1 even and 4 odd (or 4 odd and 1 even), the groups formed are: 2-5-7-9-11, 4-7-9-11-\n2nd category: 2 even and 3 odd (or 3 odd and 2 even), the groups formed are: 2-4-7-9-11, 4-6-9-11-1, 6-8-11-1-3, 8-10-1-3-5, 10-12-3-5-7, 12-2-5-7-9, totaling 6 types.\n3rd category: 3 even and 2 odd (or 2 odd and 3 even), the groups formed are: 2-4-6-9-11, 4-6-8-11-1, 6-8-10-1-3, 8-10-12-3-5, 10-12-2-5-7, 12-2-4-7-9, totaling 6 types.\n4th category: 4 even and 1 odd (or 1 odd and 4 even), the groups formed are: 2-4-6-8-11, 4-6-8-10-1, 6-8-10-12-3, 8-10-12-2-5, 10-12-2-4-7, 12-2-4-6-9, totaling 6 types. According to the calculation method: $6+6+(6+6+6+6)=6+6+24=36$ (types). Therefore: There are a total of 36 methods to form teams.", "answer": "36", "source": "olympiads", "problem_type": "Combinatorics", "uuid": "3764ec03-dc9d-5c6e-8a85-3e707f13bd9d", "difficulty_band": "medium", "correctness_count": 2}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\n8. (10 points) 12 Smurfs are sitting around a round table, each Smurf hates the 2 Smurfs sitting next to him, but does not hate the other 9 Smurfs. Papa Smurf needs to send out a team of 5 Smurfs to rescue Smurfette, who was captured by Gargamel, and the team cannot include Smurfs who hate each other. There are $\\qquad$ ways to form the team.", "candidates": [{"kind": "greedy", "answer": "1. **Total Number of Smurfs**: There are 12 Smurfs in total.\n2. **Number of Smurfs who Hate Each Other**: \n - The number of Smurfs who hate each other is given by the combination formula $\\binom{12}{2} = 66$.\n3. **Number of Smurfs who Hate Each Other**: \n - The number of Smurfs who hate each other is given by the number of ways to choose 2 Smurfs from 12, which is $\\binom{12}{2} = 66$.\n4. **Number of Smurfs who Hate Each Other and Hate Each other**: \n - The number of Smurfs who hate each other and hate each other is given by the number of ways to choose 2 Smurfs from 66, which is $\\binom{66}{2} = 2310$.\n5. **Number of Smurfs who Hate Each Other and Hate Each other and Hate each other**: \n - The number of Smurfs who hate each other and hate each other and hate each other is given by the number of ways to choose 2 Smurfs from 2310, which is $\\binom{2310}{2} = 2310$.\n6. **Number of Smurfs who Hate Each Other and Hate Each other and Hate each other and Hate each other**: \n - The number of Smurfs who hate each other and hate each other and hate each other is given by the number of ways to choose 2 Smurfs from 2310, which is $\\binom{2310}{2} = 2310$.\n7. **Number of Smurfs who Hate Each Other and Hate Each other and Hate each other and Hate each other**: \n - The number of Smurfs who hate each other and hate each other and hate each other is given by the number of ways to choose 2 Smurfs from 2310, which is $\\binom{2310}{2} = 2310$.\n8. **Number of Smurfs who Hate Each Other and Hate Each other and Hate"}, {"kind": "sampled", "answer": "The total number of Smurfs sitting around the table is \n$$4 \\times 12 = 48.$$\n\nThe number of Smurfs sitting next to each other is \n$$9 \\times 9 = 81.$$\n\nTherefore, the number of Smurfs who Boyle is notnext to is \n$$48 - 81 = -13.$$\n\nThere are $2^{\\boxed{13}}$ ways to choose 5 Smurfs for the team.\n\nFinal answer:\n#### 2^{13}"}]} +{"index": 98, "row": {"problem": "25. Anna, Bridgit and Carol run in a $100 \\mathrm{~m}$ race. When Anna finishes, Bridgit is $16 \\mathrm{~m}$ behind her and when Bridgit finishes, Carol is $25 \\mathrm{~m}$ behind her. The girls run at constant speeds throughout the race. How far behind was Carol when Anna finished?\nA $37 \\mathrm{~m}$\nB $41 \\mathrm{~m}$\nC $50 \\mathrm{~m}$\nD $55 \\mathrm{~m}$\nE $60 \\mathrm{~m}$", "solution": "25. A Carol finishes 25 metres behind Bridgit, so she travels 75 metres while Bridgit runs 100 metres. Therefore she runs 3 metres for every 4 metres Bridgit runs. When Anna finishes, Bridgit has run 84 metres, so that at that time Carol has run $\\frac{3}{4} \\times 84$ metres $=63$ metres. Hence Carol finishes $(100-63)$ metres $=37$ metres behind Anna.", "answer": "37\\mathrm{~}", "source": "olympiads", "problem_type": "Algebra", "uuid": "f7bf0e57-7e91-5810-bab6-d1e4ad0d4461", "difficulty_band": "hard", "correctness_count": 1}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\n25. Anna, Bridgit and Carol run in a $100 \\mathrm{~m}$ race. When Anna finishes, Bridgit is $16 \\mathrm{~m}$ behind her and when Bridgit finishes, Carol is $25 \\mathrm{~m}$ behind her. The girls run at constant speeds throughout the race. How far behind was Carol when Anna finished?\nA $37 \\mathrm{~m}$\nB $41 \\mathrm{~m}$\nC $50 \\mathrm{~m}$\nD $55 \\mathrm{~m}$\nE $60 \\mathrm{~m}$", "candidates": [{"kind": "greedy", "answer": "1. **Define Variables:**\n Let \\( v_A \\), \\( v_B \\), and \\( v_C \\) be the speeds of Anna, Bridgit, and Carol, respectively.\n\n2. **Set Up Equations:**\n From the problem, we have the following equations:\n \\[\n \\frac{d}{100} = \\frac{d}{v_A} + \\frac{d}{v_B} + \\frac{d}{v_C}\n \\]\n \\[\n \\frac{d}{100} = \\frac{d}{v_A} + \\frac{d}{v_B} + \\frac{d}{v_C} - \\frac{d}{v_A} - \\frac{d}{v_B} - \\frac{d}{v_C}\n \\]\n\n3. **Simplify the Equations:**\n Simplify the equations by canceling out the common terms:\n \\[\n \\frac{d}{100} = \\frac{d}{v_A} + \\frac{d}{v_B} + \\frac{d}{v_C} - \\frac{d}{v_A} - \\frac{d}{v_B} - \\frac{d}{v_C} = \\frac{d}{v_A} + \\frac{d}{v_B} + \\frac{d}{v_C} - \\frac{d}{v_A} - \\frac{d}{v_B} - \\frac{d}{v_C} = 0\n \\]\n \\[\n \\frac{d}{100} = \\frac{d}{v_A} + \\frac{d}{v_B} + \\frac{d}{v_C} - \\frac{d}{v_A} - \\frac{d}{v_B} - \\frac{d}{v_C} = 0\n \\]\n\n4. **Solve the System of Equations:**\n From the simplified equations, we have:\n \\[\n \\frac{d}{v_A} + \\frac{d}{v_B} + \\frac{d}{v_C} = \\frac{d}{100}\n \\]\n \\[\n \\frac{d}{v_A} + \\frac{d}{v"}, {"kind": "sampled", "answer": "1. Let $x$ be the distance that was run by Anna.\n2. Since Bridgit was 16 meters behind Anna, her distance was $x + 16$.\n3. Since Carol was 25 meters behind Anna, her distance was $x - 25$.\n4. The problem states that when Anna finishes, Bridova and Carol are 16 meters behind.\n5. Therefore, we can set up the following equation:\n \\[ x + 16 + (x - 25) = 100 \\]\n6. Simplify the equation:\n \\[ x + 16 + x - 25 = 100 \\Rightarrow 2x - 9 = 100 \\Rightarrow 2x = 109 \\Rightarrow x = 54.5 \\]\n7. Substitute $x = 54.5$ into the equation:\n \\[ 54.5 + 16 = 100 \\Rightarrow 60.5 = 100 \\]\n8. This equation is not consistent, so let's re-evaluate the setup.\n \n9. The key insight is to recognize the relationship between the distances when Anna and Brid revive. Since Anna covered $x$ meters, and she was 16 meters behind when Brid Ing got there, the difference is in the distances run by screen and screen.\n\n10. The race is 100 meters long and the distances run by each can be calculated by subtracting the running distances to the screen and screen. When Brid courty gets there, the distances run by screen and screen are equal.\n\nConclusion:\nThe difference in the distances run by screen and screen is 100 meters.\nThe final answer:\n#### D"}]} +{"index": 99, "row": {"problem": "Example 13 Given that $a$, $b$, and $c$ are all positive integers, and the parabola $y=a x^{2}+b x+c$ intersects the $x$-axis at two distinct points $A$ and $B$. If the distances from $A$ and $B$ to the origin are both less than 1, find the minimum value of $a+b+c$.\n(1996, National Junior High School Mathematics League)", "solution": "Solution: Let $A\\left(x_{1}, 0\\right)$ and $B\\left(x_{2}, 0\\right)$, and $x_{1} < 0 < x_{2}$, \nthen $x_{1} < 0, \\\\ \n\\therefore b > 2 \\sqrt{a c} . \\\\ \n\\text{Also, } \\because |O A| = |x_{1}| > 1$. Therefore, the parabola opens upwards, and when $x = -1$, $y > 0$, so $a(-1)^{2} + b(-1) + c > 0$, which means $b < a + c + 1$. Since $b > 2 \\sqrt{a c}$, we have $2 \\sqrt{a c} + 1 < a + c$, \ni.e., $(\\sqrt{a} - \\sqrt{c})^{2} > 1$. \nFrom equation (2), we get $\\sqrt{a} - \\sqrt{c} > 1$, \n\ni.e., $\\sqrt{a} > \\sqrt{c} + 1$. \nThus, $a > (\\sqrt{c} + 1)^{2} \\geq (\\sqrt{1} + 1)^{2} = 4$. \nSolving, we get $a \\geq 5$. \n$$\n\\begin{array}{l}\n\\text{Also, } \\because b > 2 \\sqrt{a c} \\geq 2 \\sqrt{5 \\times 1} > 4, \\\\\n\\therefore b \\geq 5 .\n\\end{array}\n$$\n\nWhen $a = 5, b = 5, c = 1$, the parabola $y = 5 x^{2} + 5 x + 1$ satisfies the given conditions. \nTherefore, the minimum value of $a + b + c$ is 11.", "answer": "11", "source": "cn_contest", "problem_type": "Algebra", "uuid": "145da507-b5fa-5f63-b6dd-5097b644c349", "difficulty_band": "medium", "correctness_count": 2}, "prompt": "Solve this mathematics problem carefully. Give concise, checkable reasoning and a clear final answer.\n\nExample 13 Given that $a$, $b$, and $c$ are all positive integers, and the parabola $y=a x^{2}+b x+c$ intersects the $x$-axis at two distinct points $A$ and $B$. If the distances from $A$ and $B$ to the origin are both less than 1, find the minimum value of $a+b+c$.\n(1996, National Junior High School Mathematics League)", "candidates": [{"kind": "greedy", "answer": "To find the minimum value of $a+b+c$, we need to consider the conditions given in the problem.\n\nSince the parabola intersects the $x$-axis at two distinct points $A$ and $B$, we know that the discriminant of the quadratic equation $ax^2+bx+c=0$ must be greater than 0.\n\nThe discriminant is given by $b^2-4ac$.\n\nFor the parabola to intersect the $x$-axis at two distinct points, the discriminant must be greater than 0.\n\nTherefore, we have the inequality:\n\\[ b^2-4ac > 0 \\]\n\nWe also know that $a$, $b$, and $c$ are all positive integers.\n\nTo minimize $a+b+c$, we want to minimize the values of $a$, $b$, and $c$.\n\nThe smallest possible values for $a$, $b$, and $c$ are 1, 1, and 1, respectively.\n\nTherefore, the minimum value of $a+b+c$ is:\n\\[ 1+1+1 = \\boxed{3} \\]\n\nFinal answer:\n#### 3"}, {"kind": "sampled", "answer": "Let the roots of the equation $ax^2+bx+c=0$ be $x_1$ and $x_2$. By Vieta's formulas, we have:\n\\[ x_1 + x_2 = -\\frac{b}{a} \\quad \\text{and} \\quad x_1 x_2 = \\frac{c}{a} \\]\n\nWe are given that the distances from $A$ and $B$ to the origin are both less than 1. We know that the distance from a point to the origin is defined as the absolute value of the difference between the two points.\n\nSo, we have:\n\\[ |x_1 - x_2| < 1 \\Rightarrow |x_1 + x_2| < 2 \\Rightarrow -2 < x_1 + x_2 < 2 \\]\n\nThis means that the two roots are such that one is closer to 0 than the other. This situation occurs when the quadratic equation $ax^2+bx+c=0$ has two real roots, one greater than 0 and the other less than 0.\n\nWe can express the roots as $-m\\cdot\\frac{b}{a}$ and $\\frac{c}{a}$, where $m$ is a positive integer.\n\nNow, we can use Vieta's formulas:\n\\[ x_1 + x_2 = -\\frac{b}{a} \\Rightarrow -m\\cdot\\frac{b}{a} + \\frac{c}{a} = -\\frac{b}{a} \\Rightarrow -m\\cdot\\frac{b}{a} = \\frac{c}{a} \\Rightarrow -m = \\frac{c}{a} \\Rightarrow m = -\\frac{c}{a} \\]\n\nSince $m$ is a positive integer, we have $a = -c$.\n\nNow, we need to find the minimum value of $a+b+c$. Substitute $a = -c$ into the expression:\n\\[ a + b + c = -c + b + c = b + 2c \\]\n\nTo minimize this expression, we need to minimize $b + 2c$. Since $b$ is a positive integer, the smallest possible value of $b$ is 1.\n\nThe"}]}