logs/step_111-b_0.log

{
  "log_file": [
    "step_111-b_0.log",
    "step_111-b_1.log",
    "step_111-b_2.log",
    "step_111-b_3.log",
    "step_111-b_4.log",
    "step_111-b_5.log"
  ],
  "query": [
    "<s>Below is an Instruction section that describes a task, paired with an Input section that provides further context.\nWrite in the Response section that appropriately completes the request.\n\n### Instruction:\nAnswer a math question in the input.\n\nTo assist you, you can invoke a math-aware search API (i.e., SEARCH) or a computation API (COMPUTE), and I will insert the returned API results for you right after each valid SEARCH or COMPUTE calls.\n\nThe SEARCH API is followed by its parameters which are a list of keywords in JSON format, for example:\n\nSEARCH[\"$x^2 = -1$\", \"imaginary numbers\"]\n\nDO NOT mix text and math in one JSON item, i.e. instead of writing:\n\nSEARCH['$what kind of curve is defined by x^2 - y^2 = 4$']\n\nwrite keyword by keyword with only one type in each:\n\nSEARCH[\"curve\", \"defined by\", \"$x^2 - y^2 = 4$\"]\n\nFor the COMPUTE API, it is also followed by its parameters in JSON. The first parameter `mode' is chosen from `calculate', `simplify' or `solve *', whereas the second parameter is the symbolic expression in LaTeX.\n\nFor example, to calculate sine of 270 degree, you can do:\n\nCOMPUTE[\"calculate\", \"\\\\sin(270 \\\\times \\\\frac{\\\\pi}{180})\"]\n\nTo simplify $\\sin^2 x + \\cos^2 x$, you can do:\n\nCOMPUTE[\"simplify\", \"\\\\sin^2(x) + \\\\cos^2(x)\"]\n\nAnd to solve $y = 1 - 2 y^2$ for y, you can do:\n\nCOMPUTE[\"solve y\", \"y = 1 - 2 y^2\"]\n\nFor the SEARCH API, only consider helpful API results for your goal, ignore irrelevant ones.\nFor the COMPUTE API, remember it is limited to simple tasks. It does not support linear algebra, nor matrix manipulations.\n\nWhen the API result is helpful, you can just rely on the result or extract the final answer from it directly, in such case, there is no need to answer from the begining and redo any existing derivations in the result.\n\nWhen API results are not helpful, ignore the results and answer the given math question directly!\n\nAt the end, indicate your final answer in boxed LaTeX. For example, if you think the final answer is \\sqrt{3}, write it as \\boxed{\\sqrt{3}} (in boxed LaTeX) at the very end of your output.\n\nTake a deep breath and now I will hand the math question to you!\n\n### Input:\nWhen the vectors $\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}$ and $\\begin{pmatrix} -1 \\\\ 3 \\end{pmatrix}$ are both projected onto the same vector $\\mathbf{v},$ the result is $\\mathbf{p}$ in both cases.  Find $\\mathbf{p}.$\n\n### Response:\n",
    "<s>Below is an Instruction section that describes a task, paired with an Input section that provides further context.\nWrite in the Response section that appropriately completes the request.\n\n### Instruction:\nAnswer a math question in the input.\n\nTo assist you, you can invoke a math-aware search API (i.e., SEARCH) or a computation API (COMPUTE), and I will insert the returned API results for you right after each valid SEARCH or COMPUTE calls.\n\nThe SEARCH API is followed by its parameters which are a list of keywords in JSON format, for example:\n\nSEARCH[\"$x^2 = -1$\", \"imaginary numbers\"]\n\nDO NOT mix text and math in one JSON item, i.e. instead of writing:\n\nSEARCH['$what kind of curve is defined by x^2 - y^2 = 4$']\n\nwrite keyword by keyword with only one type in each:\n\nSEARCH[\"curve\", \"defined by\", \"$x^2 - y^2 = 4$\"]\n\nFor the COMPUTE API, it is also followed by its parameters in JSON. The first parameter `mode' is chosen from `calculate', `simplify' or `solve *', whereas the second parameter is the symbolic expression in LaTeX.\n\nFor example, to calculate sine of 270 degree, you can do:\n\nCOMPUTE[\"calculate\", \"\\\\sin(270 \\\\times \\\\frac{\\\\pi}{180})\"]\n\nTo simplify $\\sin^2 x + \\cos^2 x$, you can do:\n\nCOMPUTE[\"simplify\", \"\\\\sin^2(x) + \\\\cos^2(x)\"]\n\nAnd to solve $y = 1 - 2 y^2$ for y, you can do:\n\nCOMPUTE[\"solve y\", \"y = 1 - 2 y^2\"]\n\nFor the SEARCH API, only consider helpful API results for your goal, ignore irrelevant ones.\nFor the COMPUTE API, remember it is limited to simple tasks. It does not support linear algebra, nor matrix manipulations.\n\nWhen the API result is helpful, you can just rely on the result or extract the final answer from it directly, in such case, there is no need to answer from the begining and redo any existing derivations in the result.\n\nWhen API results are not helpful, ignore the results and answer the given math question directly!\n\nAt the end, indicate your final answer in boxed LaTeX. For example, if you think the final answer is \\sqrt{3}, write it as \\boxed{\\sqrt{3}} (in boxed LaTeX) at the very end of your output.\n\nTake a deep breath and now I will hand the math question to you!\n\n### Input:\nWhen the vectors $\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}$ and $\\begin{pmatrix} -1 \\\\ 3 \\end{pmatrix}$ are both projected onto the same vector $\\mathbf{v},$ the result is $\\mathbf{p}$ in both cases.  Find $\\mathbf{p}.$\n\n### Response:\nSEARCH[\"\\\\mathbf{p} = \\\\mathbf{p}\"]\n\nHere are the results:\n--- RESULTS BEGIN ---\nURL: /tuna1/scratch/w32zhong/corpus/MATH/train/intermediate_algebra/2119.json\n\n#### Similar Question\nSuppose that a parabola has vertex $\\left(\\frac{1}{4},-\\frac{9}{8}\\right)$ and equation $y = ax^2 + bx + c$, where $a > 0$ and $a + b + c$ is an integer.  Find the smallest possible value of $a.$\n\n#### User Answer\nSince the vertex is at $\\left(\\frac{1}{4}, -\\frac{9}{8}\\right)$, the equation of the parabola can be expressed in the form\n\\[y=a\\left(x-\\frac{1}{4}\\right)^2-\\frac{9}{8}.\\]Expanding, we find that\n\\[y=a\\left(x^2-\\frac{x}{2}+\\frac{1}{16}\\right)-\\frac{9}{8} =ax^2-\\frac{ax}{2}+\\frac{a}{16}-\\frac{9}{8}.\\]From the problem, we know that the parabola can be expressed in the form $y=ax^2+bx+c$, where $a+b+c$ is an integer. From the above equation, we can conclude that $a=a$, $b = -\\frac{a}{2}$, and $c = \\frac{a}{16}-\\frac{9}{8}$.  Adding up all of these gives us\n\\[a + b + c = \\frac{9a-18}{16} = \\frac{9(a - 2)}{16}.\\]Let $n = a + b + c.$  Then $\\frac{9(a - 2)}{16} = n,$ so\n\\[a = \\frac{16n + 18}{9}.\\]For $a$ to be positive, we must have $16n + 18 > 0,$ or $n > -\\frac{9}{8}.$  Setting $n = -1,$ we get $a = \\frac{2}{9}.$\n\nThus, the smallest possible value of $a$ is $\\frac{2}{9}.$\n\n--- RESULTS END ---\n\n",
    "<s>Below is an Instruction section that describes a task, paired with an Input section that provides further context.\nWrite in the Response section that appropriately completes the request.\n\n### Instruction:\nAnswer a math question in the input.\n\nTo assist you, you can invoke a math-aware search API (i.e., SEARCH) or a computation API (COMPUTE), and I will insert the returned API results for you right after each valid SEARCH or COMPUTE calls.\n\nThe SEARCH API is followed by its parameters which are a list of keywords in JSON format, for example:\n\nSEARCH[\"$x^2 = -1$\", \"imaginary numbers\"]\n\nDO NOT mix text and math in one JSON item, i.e. instead of writing:\n\nSEARCH['$what kind of curve is defined by x^2 - y^2 = 4$']\n\nwrite keyword by keyword with only one type in each:\n\nSEARCH[\"curve\", \"defined by\", \"$x^2 - y^2 = 4$\"]\n\nFor the COMPUTE API, it is also followed by its parameters in JSON. The first parameter `mode' is chosen from `calculate', `simplify' or `solve *', whereas the second parameter is the symbolic expression in LaTeX.\n\nFor example, to calculate sine of 270 degree, you can do:\n\nCOMPUTE[\"calculate\", \"\\\\sin(270 \\\\times \\\\frac{\\\\pi}{180})\"]\n\nTo simplify $\\sin^2 x + \\cos^2 x$, you can do:\n\nCOMPUTE[\"simplify\", \"\\\\sin^2(x) + \\\\cos^2(x)\"]\n\nAnd to solve $y = 1 - 2 y^2$ for y, you can do:\n\nCOMPUTE[\"solve y\", \"y = 1 - 2 y^2\"]\n\nFor the SEARCH API, only consider helpful API results for your goal, ignore irrelevant ones.\nFor the COMPUTE API, remember it is limited to simple tasks. It does not support linear algebra, nor matrix manipulations.\n\nWhen the API result is helpful, you can just rely on the result or extract the final answer from it directly, in such case, there is no need to answer from the begining and redo any existing derivations in the result.\n\nWhen API results are not helpful, ignore the results and answer the given math question directly!\n\nAt the end, indicate your final answer in boxed LaTeX. For example, if you think the final answer is \\sqrt{3}, write it as \\boxed{\\sqrt{3}} (in boxed LaTeX) at the very end of your output.\n\nTake a deep breath and now I will hand the math question to you!\n\n### Input:\nWhen the vectors $\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}$ and $\\begin{pmatrix} -1 \\\\ 3 \\end{pmatrix}$ are both projected onto the same vector $\\mathbf{v},$ the result is $\\mathbf{p}$ in both cases.  Find $\\mathbf{p}.$\n\n### Response:\n",
    "<s>Below is an Instruction section that describes a task, paired with an Input section that provides further context.\nWrite in the Response section that appropriately completes the request.\n\n### Instruction:\nAnswer a math question in the input.\n\nTo assist you, you can invoke a math-aware search API (i.e., SEARCH) or a computation API (COMPUTE), and I will insert the returned API results for you right after each valid SEARCH or COMPUTE calls.\n\nThe SEARCH API is followed by its parameters which are a list of keywords in JSON format, for example:\n\nSEARCH[\"$x^2 = -1$\", \"imaginary numbers\"]\n\nDO NOT mix text and math in one JSON item, i.e. instead of writing:\n\nSEARCH['$what kind of curve is defined by x^2 - y^2 = 4$']\n\nwrite keyword by keyword with only one type in each:\n\nSEARCH[\"curve\", \"defined by\", \"$x^2 - y^2 = 4$\"]\n\nFor the COMPUTE API, it is also followed by its parameters in JSON. The first parameter `mode' is chosen from `calculate', `simplify' or `solve *', whereas the second parameter is the symbolic expression in LaTeX.\n\nFor example, to calculate sine of 270 degree, you can do:\n\nCOMPUTE[\"calculate\", \"\\\\sin(270 \\\\times \\\\frac{\\\\pi}{180})\"]\n\nTo simplify $\\sin^2 x + \\cos^2 x$, you can do:\n\nCOMPUTE[\"simplify\", \"\\\\sin^2(x) + \\\\cos^2(x)\"]\n\nAnd to solve $y = 1 - 2 y^2$ for y, you can do:\n\nCOMPUTE[\"solve y\", \"y = 1 - 2 y^2\"]\n\nFor the SEARCH API, only consider helpful API results for your goal, ignore irrelevant ones.\nFor the COMPUTE API, remember it is limited to simple tasks. It does not support linear algebra, nor matrix manipulations.\n\nWhen the API result is helpful, you can just rely on the result or extract the final answer from it directly, in such case, there is no need to answer from the begining and redo any existing derivations in the result.\n\nWhen API results are not helpful, ignore the results and answer the given math question directly!\n\nAt the end, indicate your final answer in boxed LaTeX. For example, if you think the final answer is \\sqrt{3}, write it as \\boxed{\\sqrt{3}} (in boxed LaTeX) at the very end of your output.\n\nTake a deep breath and now I will hand the math question to you!\n\n### Input:\nWhen the vectors $\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}$ and $\\begin{pmatrix} -1 \\\\ 3 \\end{pmatrix}$ are both projected onto the same vector $\\mathbf{v},$ the result is $\\mathbf{p}$ in both cases.  Find $\\mathbf{p}.$\n\n### Response:\nSEARCH[\"\\\\mathbf{p} = \\\\mathbf{p}\"]\n\nHere are the results:\n--- RESULTS BEGIN ---\nURL: /tuna1/scratch/w32zhong/corpus/MATH/train/intermediate_algebra/552.json\n\n#### Similar Question\nLet $a_1,$ $a_2,$ $a_3,$ $\\dots$ be a sequence of real numbers satisfying\n\\[a_n = a_{n - 1} a_{n + 1}\\]for all $n \\ge 2.$  If $a_1 = 1 + \\sqrt{7}$ and $a_{1776} = 13 + \\sqrt{7},$ then determine $a_{2009}.$\n\n#### User Answer\nFrom the given recursion,\n\\[a_{n + 1} = \\frac{a_n}{a_{n - 1}}.\\]Let $a = a_1$ and $b = a_2.$  Then\n\\begin{align*}\na_3 &= \\frac{a_2}{a_1} = \\frac{b}{a}, \\\\\na_4 &= \\frac{a_3}{a_2} = \\frac{b/a}{b} = \\frac{1}{a}, \\\\\na_5 &= \\frac{a_4}{a_3} = \\frac{1/a}{b/a} = \\frac{1}{b}, \\\\\na_6 &= \\frac{a_5}{a_4} = \\frac{1/b}{1/a} = \\frac{a}{b}, \\\\\na_7 &= \\frac{a_6}{a_5} = \\frac{a/b}{1/b} = a, \\\\\na_8 &= \\frac{a_7}{a_6} = \\frac{a}{a/b} = b.\n\\end{align*}Since $a_7 = a = a_1$ and $a_8 = b = a_2,$ and each term depends only on the two previous terms, the sequence is periodic from here on.  Furthermore, the length of the period is 6.  Therefore, $a_6 = a_{1776} = 13 + \\sqrt{7}$ and $a_{2009} = a_5.$  Also, $a_7 = a_1,$ and\n\\[a_7 = \\frac{a_6}{a_5}.\\]Hence,\n\\[a_5 = \\frac{a_6}{a_7} = \\frac{13 + \\sqrt{7}}{1 + \\sqrt{7}} = \\frac{(13 + \\sqrt{7})(\\sqrt{7} - 1)}{(1 + \\sqrt{7})(\\sqrt{7} - 1)} = \\frac{-6 + 12 \\sqrt{7}}{6} = -1 + 2 \\sqrt{7}.\\]\n\n--- RESULTS END ---\n\n",
    "<s>Below is an Instruction section that describes a task, paired with an Input section that provides further context.\nWrite in the Response section that appropriately completes the request.\n\n### Instruction:\nAnswer a math question in the input.\n\nTo assist you, you can invoke a math-aware search API (i.e., SEARCH) or a computation API (COMPUTE), and I will insert the returned API results for you right after each valid SEARCH or COMPUTE calls.\n\nThe SEARCH API is followed by its parameters which are a list of keywords in JSON format, for example:\n\nSEARCH[\"$x^2 = -1$\", \"imaginary numbers\"]\n\nDO NOT mix text and math in one JSON item, i.e. instead of writing:\n\nSEARCH['$what kind of curve is defined by x^2 - y^2 = 4$']\n\nwrite keyword by keyword with only one type in each:\n\nSEARCH[\"curve\", \"defined by\", \"$x^2 - y^2 = 4$\"]\n\nFor the COMPUTE API, it is also followed by its parameters in JSON. The first parameter `mode' is chosen from `calculate', `simplify' or `solve *', whereas the second parameter is the symbolic expression in LaTeX.\n\nFor example, to calculate sine of 270 degree, you can do:\n\nCOMPUTE[\"calculate\", \"\\\\sin(270 \\\\times \\\\frac{\\\\pi}{180})\"]\n\nTo simplify $\\sin^2 x + \\cos^2 x$, you can do:\n\nCOMPUTE[\"simplify\", \"\\\\sin^2(x) + \\\\cos^2(x)\"]\n\nAnd to solve $y = 1 - 2 y^2$ for y, you can do:\n\nCOMPUTE[\"solve y\", \"y = 1 - 2 y^2\"]\n\nFor the SEARCH API, only consider helpful API results for your goal, ignore irrelevant ones.\nFor the COMPUTE API, remember it is limited to simple tasks. It does not support linear algebra, nor matrix manipulations.\n\nWhen the API result is helpful, you can just rely on the result or extract the final answer from it directly, in such case, there is no need to answer from the begining and redo any existing derivations in the result.\n\nWhen API results are not helpful, ignore the results and answer the given math question directly!\n\nAt the end, indicate your final answer in boxed LaTeX. For example, if you think the final answer is \\sqrt{3}, write it as \\boxed{\\sqrt{3}} (in boxed LaTeX) at the very end of your output.\n\nTake a deep breath and now I will hand the math question to you!\n\n### Input:\nWhen the vectors $\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}$ and $\\begin{pmatrix} -1 \\\\ 3 \\end{pmatrix}$ are both projected onto the same vector $\\mathbf{v},$ the result is $\\mathbf{p}$ in both cases.  Find $\\mathbf{p}.$\n\n### Response:\n",
    "<s>Below is an Instruction section that describes a task, paired with an Input section that provides further context.\nWrite in the Response section that appropriately completes the request.\n\n### Instruction:\nAnswer a math question in the input.\n\nTo assist you, you can invoke a math-aware search API (i.e., SEARCH) or a computation API (COMPUTE), and I will insert the returned API results for you right after each valid SEARCH or COMPUTE calls.\n\nThe SEARCH API is followed by its parameters which are a list of keywords in JSON format, for example:\n\nSEARCH[\"$x^2 = -1$\", \"imaginary numbers\"]\n\nDO NOT mix text and math in one JSON item, i.e. instead of writing:\n\nSEARCH['$what kind of curve is defined by x^2 - y^2 = 4$']\n\nwrite keyword by keyword with only one type in each:\n\nSEARCH[\"curve\", \"defined by\", \"$x^2 - y^2 = 4$\"]\n\nFor the COMPUTE API, it is also followed by its parameters in JSON. The first parameter `mode' is chosen from `calculate', `simplify' or `solve *', whereas the second parameter is the symbolic expression in LaTeX.\n\nFor example, to calculate sine of 270 degree, you can do:\n\nCOMPUTE[\"calculate\", \"\\\\sin(270 \\\\times \\\\frac{\\\\pi}{180})\"]\n\nTo simplify $\\sin^2 x + \\cos^2 x$, you can do:\n\nCOMPUTE[\"simplify\", \"\\\\sin^2(x) + \\\\cos^2(x)\"]\n\nAnd to solve $y = 1 - 2 y^2$ for y, you can do:\n\nCOMPUTE[\"solve y\", \"y = 1 - 2 y^2\"]\n\nFor the SEARCH API, only consider helpful API results for your goal, ignore irrelevant ones.\nFor the COMPUTE API, remember it is limited to simple tasks. It does not support linear algebra, nor matrix manipulations.\n\nWhen the API result is helpful, you can just rely on the result or extract the final answer from it directly, in such case, there is no need to answer from the begining and redo any existing derivations in the result.\n\nWhen API results are not helpful, ignore the results and answer the given math question directly!\n\nAt the end, indicate your final answer in boxed LaTeX. For example, if you think the final answer is \\sqrt{3}, write it as \\boxed{\\sqrt{3}} (in boxed LaTeX) at the very end of your output.\n\nTake a deep breath and now I will hand the math question to you!\n\n### Input:\nWhen the vectors $\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}$ and $\\begin{pmatrix} -1 \\\\ 3 \\end{pmatrix}$ are both projected onto the same vector $\\mathbf{v},$ the result is $\\mathbf{p}$ in both cases.  Find $\\mathbf{p}.$\n\n### Response:\nSEARCH[\"\\\\mathbf{p} = \\\\mathbf{p}\"]\n\nHere are the results:\n--- RESULTS BEGIN ---\nURL: /tuna1/scratch/w32zhong/corpus/MATH/train/intermediate_algebra/1213.json\n\n#### Similar Question\nLet\n$$p(x,y) = a_0 + a_1x + a_2y + a_3x^2 + a_4xy + a_5y^2 + a_6x^3 + a_7x^2y + a_8xy^2 + a_9y^3.$$Suppose that\n\\begin{align*}\np(0,0) &=p(1,0) = p( - 1,0) = p(0,1) = p(0, - 1)= p(1,1) = p(1, - 1) = p(2,2) = 0.\n\\end{align*}There is a point $(r,s)$ for which $p(r,s) = 0$ for all such polynomials, where $r$ and $s$ are not integers.  Find the point $(r,s).$\n\n#### User Answer\nObserve that  \\begin{align*}\np(0,0) &= a_0 = 0\\\\\np(1,0) &= a_0 + a_1 + a_3 + a_6 = a_1 + a_3 + a_6 = 0\\\\\np(-1,0) &= -a_1 + a_3 - a_6 = 0.\n\\end{align*}Adding the above two equations gives $a_3 = 0$, and so we can deduce that $a_6 = -a_1$. Similarly, plugging in $(0,1)$ and $(0,-1)$ gives $a_5 = 0$ and $a_9 = -a_2$. Now, \\begin{align*}\np(1,1) &= a_0 + a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + a_7 + a_8 + a_9\\\\\n&= 0 + a_1 + a_2 + 0 + a_4 + 0 - a_1 + a_7 + a_8 - a_2 = a_4 + a_7 + a_8 = 0\\\\\np(1,-1) &= a_0 + a_1 - a_2 + 0 - a_4 + 0 - a_1 - a_7 + a_8 + a_2\\\\ &= -a_4 - a_7 + a_8 = 0\n\\end{align*}Therefore, $a_8 = 0$ and $a_7 = -a_4$.  Finally, $$p(2,2) = 0 + 2a_1 + 2a_2 + 0 + 4a_4 + 0 - 8a_1 - 8a_4 +0 - 8a_2 = -6 a_1 - 6 a_2 - 4 a_4 = 0.$$Hence, $3a_1 + 3a_2 + 2a_4 = 0$. Now,  \\begin{align*}\np(x,y) &= 0 + a_1x + a_2y + 0 + a_4xy + 0 - a_1x^3 - a_4x^2y + 0 - a_2y^3\\\\\n&= a_1 x(1-x)(1+x) + a_2 y(1-y)(1+y) + xy (1-x) a_4 \\\\\n&= a_1 x(1 - x)(1 + x) + a_2 y(1 - y)(1 + y) - \\left( \\frac{3}{2} a_1 + \\frac{3}{2} a_2 \\right) xy(1 - x) \\\\\n&= a_1 \\left( x - x^3 - \\frac{3}{2} xy(1 - x) \\right) + a_2 \\left( y - y^3 - \\frac{3}{2} xy(1 - x) \\right).\n\\end{align*}If $p(r,s) = 0$ for every such polynomial, then\n\\begin{align*}\nr - r^3 - \\frac{3}{2} rs (1 - r) &= 0, \\\\\ns - s^3 - \\frac{3}{2} rs (1 - r) &= 0.\n\\end{align*}These factor as\n\\begin{align*}\n\\frac{1}{2} r(1 - r)(2r - 3s + 2) &= 0, \\\\\n\\frac{1}{2} s(3r^2 - 3r - 2s^2 + 2) &= 0.\n\\end{align*}Hence, $r = 0,$ $r = 1,$ or $r = \\frac{3s - 2}{2}.$\n\nSubstituting $r = 0$ into the second equation, we get $s^3 = s,$ so $s = -1,$ 0, or 1.\n\nSubstituting $r = 1$ into the second equation, we again get $s^3 = s,$ so $s = -1,$ 0, or 1.\n\nSubstituting...\n\n--- RESULTS END ---\n\n"
  ],
  "response": [
    " SEARCH[\"\\\\mathbf{p} = \\\\mathbf{p}\"]</s>",
    " The result looks irrelevant, I will completely ignore it and answer the question directly.\n\nLet $\\mathbf{p} = \\begin{pmatrix} x \\\\ y \\end{pmatrix}.$  Then the given information tells us that\n\\[\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix} - \\mathbf{p} = \\begin{pmatrix} -1 \\\\ 3 \\end{pmatrix} - \\mathbf{p} = \\begin{pmatrix} 3 \\\\ -4 \\end{pmatrix} = \\mathbf{p}.\\]Then $x = -1$ and $y = 3,$ so $\\mathbf{p} = \\boxed{\\begin{pmatrix} -1 \\\\ 3 \\end{pmatrix}}.$. The answer is \\begin{pmatrix}-1\\3\\end{pmatrix}</s>",
    " SEARCH[\"\\\\mathbf{p} = \\\\mathbf{p}\"]</s>",
    " The result looks irrelevant, I will completely ignore it and answer the question directly.\n\nThe vector $\\mathbf{v}$ is the projection of $\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}$ onto $\\begin{pmatrix} -1 \\\\ 3 \\end{pmatrix}$ is\n\\[\\frac{\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix} \\cdot \\begin{pmatrix} -1 \\\\ 3 \\end{pmatrix}}{\\begin{pmatrix} -1 \\\\ 3 \\end{pmatrix} \\cdot \\begin{pmatrix} -1 \\\\ 3 \\end{pmatrix}} \\begin{pmatrix} -1 \\\\ 3 \\end{pmatrix} = \\frac{13}{\\boxed{10}} \\begin{pmatrix} -1 \\\\ 3 \\end{pmatrix} = \\boxed{\\begin{pmatrix} -13/10 \\\\ 33/10 \\end{pmatrix}}.\\]. The answer is \\begin{pmatrix}-13/10\\33/10\\end{pmatrix}</s>",
    " SEARCH[\"\\\\mathbf{p} = \\\\mathbf{p}\"]</s>",
    "The result looks irrelevant, I will completely ignore it and answer the question directly.\n\nThe vector $\\mathbf{v}$ is the projection of $\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}$ onto $\\begin{pmatrix} -1 \\\\ 3 \\end{pmatrix}$ if and only if it is a scalar multiple of $\\begin{"
  ],
  "src_path": [
    "train/precalculus/497.json",
    "train/precalculus/497.json",
    "train/precalculus/497.json",
    "train/precalculus/497.json",
    "train/precalculus/497.json",
    "train/precalculus/497.json"
  ],
  "input": [
    "When the vectors $\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}$ and $\\begin{pmatrix} -1 \\\\ 3 \\end{pmatrix}$ are both projected onto the same vector $\\mathbf{v},$ the result is $\\mathbf{p}$ in both cases.  Find $\\mathbf{p}.$",
    "When the vectors $\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}$ and $\\begin{pmatrix} -1 \\\\ 3 \\end{pmatrix}$ are both projected onto the same vector $\\mathbf{v},$ the result is $\\mathbf{p}$ in both cases.  Find $\\mathbf{p}.$",
    "When the vectors $\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}$ and $\\begin{pmatrix} -1 \\\\ 3 \\end{pmatrix}$ are both projected onto the same vector $\\mathbf{v},$ the result is $\\mathbf{p}$ in both cases.  Find $\\mathbf{p}.$",
    "When the vectors $\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}$ and $\\begin{pmatrix} -1 \\\\ 3 \\end{pmatrix}$ are both projected onto the same vector $\\mathbf{v},$ the result is $\\mathbf{p}$ in both cases.  Find $\\mathbf{p}.$",
    "When the vectors $\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}$ and $\\begin{pmatrix} -1 \\\\ 3 \\end{pmatrix}$ are both projected onto the same vector $\\mathbf{v},$ the result is $\\mathbf{p}$ in both cases.  Find $\\mathbf{p}.$",
    "When the vectors $\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}$ and $\\begin{pmatrix} -1 \\\\ 3 \\end{pmatrix}$ are both projected onto the same vector $\\mathbf{v},$ the result is $\\mathbf{p}$ in both cases.  Find $\\mathbf{p}.$"
  ],
  "output": [
    "Note that the vector $\\mathbf{p}$ must lie on the line passing through $\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}$ and $\\begin{pmatrix} -1 \\\\ 3 \\end{pmatrix}.$  This line can be parameterized by\n\\[\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix} + t \\left( \\begin{pmatrix} -1 \\\\ 3 \\end{pmatrix} - \\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix} \\right) = \\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix} + t \\begin{pmatrix} -5 \\\\ 2 \\end{pmatrix} = \\begin{pmatrix} -5t + 4 \\\\ 2t + 1 \\end{pmatrix}.\\][asy]\nusepackage(\"amsmath\");\n\nunitsize(1 cm);\n\npair A, B, O, P;\n\nA = (4,1);\nB = (-1,3);\nO = (0,0);\nP = (O + reflect(A,B)*(O))/2;\n\ndraw((-2,0)--(5,0));\ndraw((0,-1)--(0,4));\ndraw(O--A,Arrow(6));\ndraw(O--B,Arrow(6));\ndraw(O--P,Arrow(6));\ndraw(interp(A,B,-0.1)--interp(A,B,1.1),dashed);\n\nlabel(\"$\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}$\", A, N);\nlabel(\"$\\begin{pmatrix} -1 \\\\ 3 \\end{pmatrix}$\", B, N);\nlabel(\"$\\mathbf{p}$\", P, N);\n[/asy]\n\nThe vector $\\mathbf{p}$ itself will be orthogonal to the direction vector $\\begin{pmatrix} -5 \\\\ 2 \\end{pmatrix},$ so\n\\[\\begin{pmatrix} -5t + 4 \\\\ 2t + 1 \\end{pmatrix} \\cdot \\begin{pmatrix} -5 \\\\ 2 \\end{pmatrix} = 0.\\]Hence, $(-5t + 4)(-5) + (2t + 1)(2) = 0.$  Solving, we find $t = \\frac{18}{29}.$  Hence, $\\mathbf{p} = \\boxed{\\begin{pmatrix} 26/29 \\\\ 65/29 \\end{pmatrix}}.$",
    "Note that the vector $\\mathbf{p}$ must lie on the line passing through $\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}$ and $\\begin{pmatrix} -1 \\\\ 3 \\end{pmatrix}.$  This line can be parameterized by\n\\[\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix} + t \\left( \\begin{pmatrix} -1 \\\\ 3 \\end{pmatrix} - \\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix} \\right) = \\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix} + t \\begin{pmatrix} -5 \\\\ 2 \\end{pmatrix} = \\begin{pmatrix} -5t + 4 \\\\ 2t + 1 \\end{pmatrix}.\\][asy]\nusepackage(\"amsmath\");\n\nunitsize(1 cm);\n\npair A, B, O, P;\n\nA = (4,1);\nB = (-1,3);\nO = (0,0);\nP = (O + reflect(A,B)*(O))/2;\n\ndraw((-2,0)--(5,0));\ndraw((0,-1)--(0,4));\ndraw(O--A,Arrow(6));\ndraw(O--B,Arrow(6));\ndraw(O--P,Arrow(6));\ndraw(interp(A,B,-0.1)--interp(A,B,1.1),dashed);\n\nlabel(\"$\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}$\", A, N);\nlabel(\"$\\begin{pmatrix} -1 \\\\ 3 \\end{pmatrix}$\", B, N);\nlabel(\"$\\mathbf{p}$\", P, N);\n[/asy]\n\nThe vector $\\mathbf{p}$ itself will be orthogonal to the direction vector $\\begin{pmatrix} -5 \\\\ 2 \\end{pmatrix},$ so\n\\[\\begin{pmatrix} -5t + 4 \\\\ 2t + 1 \\end{pmatrix} \\cdot \\begin{pmatrix} -5 \\\\ 2 \\end{pmatrix} = 0.\\]Hence, $(-5t + 4)(-5) + (2t + 1)(2) = 0.$  Solving, we find $t = \\frac{18}{29}.$  Hence, $\\mathbf{p} = \\boxed{\\begin{pmatrix} 26/29 \\\\ 65/29 \\end{pmatrix}}.$",
    "Note that the vector $\\mathbf{p}$ must lie on the line passing through $\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}$ and $\\begin{pmatrix} -1 \\\\ 3 \\end{pmatrix}.$  This line can be parameterized by\n\\[\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix} + t \\left( \\begin{pmatrix} -1 \\\\ 3 \\end{pmatrix} - \\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix} \\right) = \\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix} + t \\begin{pmatrix} -5 \\\\ 2 \\end{pmatrix} = \\begin{pmatrix} -5t + 4 \\\\ 2t + 1 \\end{pmatrix}.\\][asy]\nusepackage(\"amsmath\");\n\nunitsize(1 cm);\n\npair A, B, O, P;\n\nA = (4,1);\nB = (-1,3);\nO = (0,0);\nP = (O + reflect(A,B)*(O))/2;\n\ndraw((-2,0)--(5,0));\ndraw((0,-1)--(0,4));\ndraw(O--A,Arrow(6));\ndraw(O--B,Arrow(6));\ndraw(O--P,Arrow(6));\ndraw(interp(A,B,-0.1)--interp(A,B,1.1),dashed);\n\nlabel(\"$\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}$\", A, N);\nlabel(\"$\\begin{pmatrix} -1 \\\\ 3 \\end{pmatrix}$\", B, N);\nlabel(\"$\\mathbf{p}$\", P, N);\n[/asy]\n\nThe vector $\\mathbf{p}$ itself will be orthogonal to the direction vector $\\begin{pmatrix} -5 \\\\ 2 \\end{pmatrix},$ so\n\\[\\begin{pmatrix} -5t + 4 \\\\ 2t + 1 \\end{pmatrix} \\cdot \\begin{pmatrix} -5 \\\\ 2 \\end{pmatrix} = 0.\\]Hence, $(-5t + 4)(-5) + (2t + 1)(2) = 0.$  Solving, we find $t = \\frac{18}{29}.$  Hence, $\\mathbf{p} = \\boxed{\\begin{pmatrix} 26/29 \\\\ 65/29 \\end{pmatrix}}.$",
    "Note that the vector $\\mathbf{p}$ must lie on the line passing through $\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}$ and $\\begin{pmatrix} -1 \\\\ 3 \\end{pmatrix}.$  This line can be parameterized by\n\\[\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix} + t \\left( \\begin{pmatrix} -1 \\\\ 3 \\end{pmatrix} - \\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix} \\right) = \\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix} + t \\begin{pmatrix} -5 \\\\ 2 \\end{pmatrix} = \\begin{pmatrix} -5t + 4 \\\\ 2t + 1 \\end{pmatrix}.\\][asy]\nusepackage(\"amsmath\");\n\nunitsize(1 cm);\n\npair A, B, O, P;\n\nA = (4,1);\nB = (-1,3);\nO = (0,0);\nP = (O + reflect(A,B)*(O))/2;\n\ndraw((-2,0)--(5,0));\ndraw((0,-1)--(0,4));\ndraw(O--A,Arrow(6));\ndraw(O--B,Arrow(6));\ndraw(O--P,Arrow(6));\ndraw(interp(A,B,-0.1)--interp(A,B,1.1),dashed);\n\nlabel(\"$\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}$\", A, N);\nlabel(\"$\\begin{pmatrix} -1 \\\\ 3 \\end{pmatrix}$\", B, N);\nlabel(\"$\\mathbf{p}$\", P, N);\n[/asy]\n\nThe vector $\\mathbf{p}$ itself will be orthogonal to the direction vector $\\begin{pmatrix} -5 \\\\ 2 \\end{pmatrix},$ so\n\\[\\begin{pmatrix} -5t + 4 \\\\ 2t + 1 \\end{pmatrix} \\cdot \\begin{pmatrix} -5 \\\\ 2 \\end{pmatrix} = 0.\\]Hence, $(-5t + 4)(-5) + (2t + 1)(2) = 0.$  Solving, we find $t = \\frac{18}{29}.$  Hence, $\\mathbf{p} = \\boxed{\\begin{pmatrix} 26/29 \\\\ 65/29 \\end{pmatrix}}.$",
    "Note that the vector $\\mathbf{p}$ must lie on the line passing through $\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}$ and $\\begin{pmatrix} -1 \\\\ 3 \\end{pmatrix}.$  This line can be parameterized by\n\\[\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix} + t \\left( \\begin{pmatrix} -1 \\\\ 3 \\end{pmatrix} - \\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix} \\right) = \\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix} + t \\begin{pmatrix} -5 \\\\ 2 \\end{pmatrix} = \\begin{pmatrix} -5t + 4 \\\\ 2t + 1 \\end{pmatrix}.\\][asy]\nusepackage(\"amsmath\");\n\nunitsize(1 cm);\n\npair A, B, O, P;\n\nA = (4,1);\nB = (-1,3);\nO = (0,0);\nP = (O + reflect(A,B)*(O))/2;\n\ndraw((-2,0)--(5,0));\ndraw((0,-1)--(0,4));\ndraw(O--A,Arrow(6));\ndraw(O--B,Arrow(6));\ndraw(O--P,Arrow(6));\ndraw(interp(A,B,-0.1)--interp(A,B,1.1),dashed);\n\nlabel(\"$\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}$\", A, N);\nlabel(\"$\\begin{pmatrix} -1 \\\\ 3 \\end{pmatrix}$\", B, N);\nlabel(\"$\\mathbf{p}$\", P, N);\n[/asy]\n\nThe vector $\\mathbf{p}$ itself will be orthogonal to the direction vector $\\begin{pmatrix} -5 \\\\ 2 \\end{pmatrix},$ so\n\\[\\begin{pmatrix} -5t + 4 \\\\ 2t + 1 \\end{pmatrix} \\cdot \\begin{pmatrix} -5 \\\\ 2 \\end{pmatrix} = 0.\\]Hence, $(-5t + 4)(-5) + (2t + 1)(2) = 0.$  Solving, we find $t = \\frac{18}{29}.$  Hence, $\\mathbf{p} = \\boxed{\\begin{pmatrix} 26/29 \\\\ 65/29 \\end{pmatrix}}.$",
    "Note that the vector $\\mathbf{p}$ must lie on the line passing through $\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}$ and $\\begin{pmatrix} -1 \\\\ 3 \\end{pmatrix}.$  This line can be parameterized by\n\\[\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix} + t \\left( \\begin{pmatrix} -1 \\\\ 3 \\end{pmatrix} - \\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix} \\right) = \\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix} + t \\begin{pmatrix} -5 \\\\ 2 \\end{pmatrix} = \\begin{pmatrix} -5t + 4 \\\\ 2t + 1 \\end{pmatrix}.\\][asy]\nusepackage(\"amsmath\");\n\nunitsize(1 cm);\n\npair A, B, O, P;\n\nA = (4,1);\nB = (-1,3);\nO = (0,0);\nP = (O + reflect(A,B)*(O))/2;\n\ndraw((-2,0)--(5,0));\ndraw((0,-1)--(0,4));\ndraw(O--A,Arrow(6));\ndraw(O--B,Arrow(6));\ndraw(O--P,Arrow(6));\ndraw(interp(A,B,-0.1)--interp(A,B,1.1),dashed);\n\nlabel(\"$\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}$\", A, N);\nlabel(\"$\\begin{pmatrix} -1 \\\\ 3 \\end{pmatrix}$\", B, N);\nlabel(\"$\\mathbf{p}$\", P, N);\n[/asy]\n\nThe vector $\\mathbf{p}$ itself will be orthogonal to the direction vector $\\begin{pmatrix} -5 \\\\ 2 \\end{pmatrix},$ so\n\\[\\begin{pmatrix} -5t + 4 \\\\ 2t + 1 \\end{pmatrix} \\cdot \\begin{pmatrix} -5 \\\\ 2 \\end{pmatrix} = 0.\\]Hence, $(-5t + 4)(-5) + (2t + 1)(2) = 0.$  Solving, we find $t = \\frac{18}{29}.$  Hence, $\\mathbf{p} = \\boxed{\\begin{pmatrix} 26/29 \\\\ 65/29 \\end{pmatrix}}.$"
  ],
  "rewards": [
    0.0,
    0.0,
    0.0,
    0.0,
    0.0,
    0.0
  ]
}