diff --git a/candidates/budgie-alignment-v2/onpolicy-error-targeted-v4/checkpoints/v4_onpolicy_rollouts_000_050.jsonl b/candidates/budgie-alignment-v2/onpolicy-error-targeted-v4/checkpoints/v4_onpolicy_rollouts_000_050.jsonl new file mode 100644 index 0000000000000000000000000000000000000000..a02a61134958cc511e8318d4b191e33080d4be20 --- /dev/null +++ b/candidates/budgie-alignment-v2/onpolicy-error-targeted-v4/checkpoints/v4_onpolicy_rollouts_000_050.jsonl @@ -0,0 +1,50 @@ +{"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"}]}