{ "log_file": [ "step_135-b_0.log", "step_135-b_1.log", "step_135-b_2.log", "step_135-b_3.log", "step_135-b_4.log", "step_135-b_5.log" ], "query": [ "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:\nThe nine points of this grid are equally spaced horizontally and vertically. The distance between two neighboring points is 1 unit. What is the area, in square units, of the region where the two triangles overlap?\n\n[asy]\nsize(80);\ndot((0,0)); dot((0,1));dot((0,2));dot((1,0));dot((1,1));dot((1,2));dot((2,0));dot((2,1));dot((2,2));\ndraw((0,0)--(2,1)--(1,2)--cycle, linewidth(0.6));\ndraw((2,2)--(0,1)--(1,0)--cycle, linewidth(0.6));\n[/asy]\n\n### Response:\n", "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:\nThe nine points of this grid are equally spaced horizontally and vertically. The distance between two neighboring points is 1 unit. What is the area, in square units, of the region where the two triangles overlap?\n\n[asy]\nsize(80);\ndot((0,0)); dot((0,1));dot((0,2));dot((1,0));dot((1,1));dot((1,2));dot((2,0));dot((2,1));dot((2,2));\ndraw((0,0)--(2,1)--(1,2)--cycle, linewidth(0.6));\ndraw((2,2)--(0,1)--(1,0)--cycle, linewidth(0.6));\n[/asy]\n\n### Response:\nSEARCH[\"\\\\cos(1)\"]\n\nHere are the results:\n--- RESULTS BEGIN ---\nURL: /tuna1/scratch/w32zhong/corpus/MATH/train/geometry/6016.json\n\n#### Similar Question\nTwelve congruent disks are placed on a circle $C$ of radius 1 in such a way that the twelve disks cover $C$, no two of the disks overlap, and so that each of the twelve disks is tangent to its two neighbors. The resulting arrangement of disks is shown in the figure below. The sum of the areas of the twelve disks can be written in the from $\\pi(a-b\\sqrt{c})$, where $a,b,c$ are positive integers and $c$ is not divisible by the square of any prime. Find $a+b+c$.\n[asy] unitsize(100); draw(Circle((0,0),1)); dot((0,0)); draw((0,0)--(1,0)); label(\"$1$\", (0.5,0), S); for (int i=0; i<12; ++i) { dot((cos(i*pi/6), sin(i*pi/6))); } for (int a=1; a<24; a+=2) { dot(((1/cos(pi/12))*cos(a*pi/12), (1/cos(pi/12))*sin(a*pi/12))); draw(((1/cos(pi/12))*cos(a*pi/12), (1/cos(pi/12))*sin(a*pi/12))--((1/cos(pi/12))*cos((a+2)*pi/12), (1/cos(pi/12))*sin((a+2)*pi/12))); draw(Circle(((1/cos(pi/12))*cos(a*pi/12), (1/cos(pi/12))*sin(a*pi/12)), tan(pi/12))); }[/asy]\n\n#### User Answer\nWe wish to find the radius of one circle, so that we can find the total area.\nNotice that for them to contain the entire circle, each pair of circles must be tangent on the larger circle. Now consider two adjacent smaller circles. This means that the line connecting the radii is a segment of length $2r$ that is tangent to the larger circle at the midpoint of the two centers. Thus, we have essentially have a regular dodecagon whose vertices are the centers of the smaller triangles circumscribed about a circle of radius $1$.\nWe thus know that the apothem of the dodecagon is equal to $1$. To find the side length, we make a triangle consisting of a vertex, the midpoint of a side, and the center of the dodecagon, which we denote $A, M,$ and $O$ respectively. Notice that $OM=1$, and that $\\triangle OMA$ is a right triangle with hypotenuse $OA$ and $m \\angle MOA = 15^\\circ$. Thus $AM = (1) \\tan{15^\\circ} = 2 - \\sqrt {3}$, which is the radius of one of the circles. The area of one circle is thus $\\pi(2 - \\sqrt {3})^{2} = \\pi (7 - 4 \\sqrt {3})$, so the area of all $12$ circles...\n\n--- RESULTS END ---\n\n", "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:\nThe nine points of this grid are equally spaced horizontally and vertically. The distance between two neighboring points is 1 unit. What is the area, in square units, of the region where the two triangles overlap?\n\n[asy]\nsize(80);\ndot((0,0)); dot((0,1));dot((0,2));dot((1,0));dot((1,1));dot((1,2));dot((2,0));dot((2,1));dot((2,2));\ndraw((0,0)--(2,1)--(1,2)--cycle, linewidth(0.6));\ndraw((2,2)--(0,1)--(1,0)--cycle, linewidth(0.6));\n[/asy]\n\n### Response:\n", "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:\nThe nine points of this grid are equally spaced horizontally and vertically. The distance between two neighboring points is 1 unit. What is the area, in square units, of the region where the two triangles overlap?\n\n[asy]\nsize(80);\ndot((0,0)); dot((0,1));dot((0,2));dot((1,0));dot((1,1));dot((1,2));dot((2,0));dot((2,1));dot((2,2));\ndraw((0,0)--(2,1)--(1,2)--cycle, linewidth(0.6));\ndraw((2,2)--(0,1)--(1,0)--cycle, linewidth(0.6));\n[/asy]\n\n### Response:\nSEARCH[\"\\\\cos(1)\"]\n\nHere are the results:\n--- RESULTS BEGIN ---\nURL: Rectangle $ABCD$ and a semicircle with diameter $AB$ are coplanar and have nonoverlapping interiors. Let $\\mathcal{R}$ denote the region enclosed by the semicircle and the rectangle. Line $\\ell$ meets the semicircle, segment $AB$, and segment $CD$ at distinct points $N$, $U$, and $T$, respectively. Line $\\ell$ divides region $\\mathcal{R}$ into two regions with areas in the ratio $1: 2$. Suppose that $AU = 84$, $AN = 126$, and $UB = 168$. Then $DA$ can be represented as $m\\sqrt {n}$, where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m + n$.\n[asy] import graph; defaultpen(linewidth(0.7)+fontsize(10)); size(500); pen zzttqq = rgb(0.6,0.2,0); pen xdxdff = rgb(0.4902,0.4902,1); /* segments and figures */ draw((0,-154.31785)--(0,0)); draw((0,0)--(252,0)); draw((0,0)--(126,0),zzttqq); draw((126,0)--(63,109.1192),zzttqq); draw((63,109.1192)--(0,0),zzttqq); draw((-71.4052,(+9166.01287-109.1192*-71.4052)/21)--(504.60925,(+9166.01287-109.1192*504.60925)/21)); draw((0,-154.31785)--(252,-154.31785)); draw((252,-154.31785)--(252,0)); draw((0,0)--(84,0)); draw((84,0)--(252,0)); draw((63,109.1192)--(63,0)); draw((84,0)--(84,-154.31785)); draw(arc((126,0),126,0,180)); /* points and labels */ dot((0,0)); label(\"$A$\",(-16.43287,-9.3374),NE/2); dot((252,0)); label(\"$B$\",(255.242,5.00321),NE/2); dot((0,-154.31785)); label(\"$D$\",(3.48464,-149.55669),NE/2); dot((252,-154.31785)); label(\"$C$\",(255.242,-149.55669),NE/2); dot((126,0)); label(\"$O$\",(129.36332,5.00321),NE/2); dot((63,109.1192)); label(\"$N$\",(44.91307,108.57427),NE/2); label(\"$126$\",(28.18236,40.85473),NE/2); dot((84,0)); label(\"$U$\",(87.13819,5.00321),NE/2); dot((113.69848,-154.31785)); label(\"$T$\",(116.61611,-149.55669),NE/2); dot((63,0)); label(\"$N'$\",(66.42398,5.00321),NE/2); label(\"$84$\",(41.72627,-12.5242),NE/2); label(\"$168$\",(167.60494,-12.5242),NE/2); dot((84,-154.31785)); label(\"$T'$\",(87.13819,-149.55669),NE/2); dot((252,0)); label(\"$I$\",(255.242,5.00321),NE/2); clip((-71.4052,-225.24323)--(-71.4052,171.51361...\n\nthe center of the semicircl is also the midpoint of $AB$. let thi point be O . let $h$ be the length of $AD$. rescal everyth by 42, so $AU = 2, AN = 3,", "Below is an Instruction section that describes a task, paired with an Input section that provides further context.\nWrite in the Response section that appropriately completes the request.\n\n### Instruction:\nAnswer a math question in the input.\n\nTo assist you, you can invoke a math-aware search API (i.e., SEARCH) or a computation API (COMPUTE), and I will insert the returned API results for you right after each valid SEARCH or COMPUTE calls.\n\nThe SEARCH API is followed by its parameters which are a list of keywords in JSON format, for example:\n\nSEARCH[\"$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:\nThe nine points of this grid are equally spaced horizontally and vertically. The distance between two neighboring points is 1 unit. What is the area, in square units, of the region where the two triangles overlap?\n\n[asy]\nsize(80);\ndot((0,0)); dot((0,1));dot((0,2));dot((1,0));dot((1,1));dot((1,2));dot((2,0));dot((2,1));dot((2,2));\ndraw((0,0)--(2,1)--(1,2)--cycle, linewidth(0.6));\ndraw((2,2)--(0,1)--(1,0)--cycle, linewidth(0.6));\n[/asy]\n\n### Response:\n", "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:\nThe nine points of this grid are equally spaced horizontally and vertically. The distance between two neighboring points is 1 unit. What is the area, in square units, of the region where the two triangles overlap?\n\n[asy]\nsize(80);\ndot((0,0)); dot((0,1));dot((0,2));dot((1,0));dot((1,1));dot((1,2));dot((2,0));dot((2,1));dot((2,2));\ndraw((0,0)--(2,1)--(1,2)--cycle, linewidth(0.6));\ndraw((2,2)--(0,1)--(1,0)--cycle, linewidth(0.6));\n[/asy]\n\n### Response:\nSEARCH[\"\\\\cos(1)\"]\n\nHere are the results:\n--- RESULTS BEGIN ---\nURL: /tuna1/scratch/w32zhong/corpus/MATH/train/geometry/6139.json\n\n#### Similar Question\nIn $\\triangle{ABC}, AB=10, \\angle{A}=30^\\circ$ , and $\\angle{C=45^\\circ}$. Let $H, D,$ and $M$ be points on the line $BC$ such that $AH\\perp{BC}$, $\\angle{BAD}=\\angle{CAD}$, and $BM=CM$. Point $N$ is the midpoint of the segment $HM$, and point $P$ is on ray $AD$ such that $PN\\perp{BC}$. Then $AP^2=\\dfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.\n\n#### User Answer\n[asy] unitsize(20); pair A = MP(\"A\",(-5sqrt(3),0)), B = MP(\"B\",(0,5),N), C = MP(\"C\",(5,0)), M = D(MP(\"M\",0.5(B+C),NE)), D = MP(\"D\",IP(L(A,incenter(A,B,C),0,2),B--C),N), H = MP(\"H\",foot(A,B,C),N), N = MP(\"N\",0.5(H+M),NE), P = MP(\"P\",IP(A--D,L(N,N-(1,1),0,10))); D(A--B--C--cycle); D(B--H--A,blue+dashed); D(A--D); D(P--N); markscalefactor = 0.05; D(rightanglemark(A,H,B)); D(rightanglemark(P,N,D)); MP(\"10\",0.5(A+B)-(-0.1,0.1),NW); [/asy]\nLet us just drop the perpendicular from $B$ to $AC$ and label the point of intersection $O$. We will use this point later in the problem. As we can see,\n$M$ is the midpoint of $BC$ and $N$ is the midpoint of $HM$\n$AHC$ is a $45-45-90$ triangle, so $\\angle{HAB}=15^\\circ$.\n$AHD$ is $30-60-90$ triangle.\n$AH$ and $PN$ are parallel lines so $PND$ is $30-60-90$ triangle also.\nThen if we use those informations we get $AD=2HD$ and\n$PD=2ND$ and $AP=AD-PD=2HD-2ND=2HN$ or $AP=2HN=HM$\nNow we know that $HM=AP$, we can find for $HM$ which is simpler to find.\nWe can use point $B$ to split it up as $HM=HB+BM$,\nWe can chase those lengths and we would get\n$AB=10$, so $OB=5$, so $BC=5\\sqrt{2}$, so $BM=\\dfrac{1}{2} \\cdot BC=\\dfrac{5\\sqrt{2}}{2}$\nWe can also use Law of Sines:\n\\[\\frac{BC}{AB}=\\frac{\\sin\\angle A}{\\sin\\angle C}\\]\\[\\frac{BC}{10}=\\frac{\\frac{1}{2}}{\\frac{\\sqrt{2}}{2}}\\implies BC=5\\sqrt{2}\\]\nThen using right triangle $AHB$, we have $HB=10 \\sin 15^\\circ$\nSo $HB=10 \\sin 15^\\circ=\\dfrac{5(\\sqrt{6}-\\sqrt{2})}{2}$.\nAnd we know that $AP = HM = HB + BM = \\frac{5(\\sqrt6-\\sqrt2)}{2} + \\frac{5\\sqrt2}{2} = \\frac{5\\sqrt6}{2}$.\nFinally if we calculate $(AP)^2$.\n$(AP)^2=\\dfrac{150}{4}=\\dfrac{75}{2}$. So our final answ...\n\n--- RESULTS END ---\n\n" ], "response": [ " SEARCH[\"\\\\cos(1)\"]", " The result looks irrelevant, I will completely ignore it and answer the question directly.\n\nThe two triangles have the same area, because they have the same height and base lengths. The height is 2, and the base lengths are 1 and 3. Therefore, the area of the triangle is $\\frac{1 \\cdot 3}{2} = \\boxed{\\frac{3}{2}}$ square units.", " SEARCH[\"\\\\cos(1)\"]", "A", " SEARCH[\"\\\\cos(1)\"]", " The result looks irrelevant, I will completely ignore it and answer the question directly.\n\nThe two triangles both have height 2 and base 4, so their combined area is $\\frac{1}{2}\\cdot4\\cdot2+ \\frac{1}{2}\\cdot4\\cdot2 = 8+8 =\\boxed{16}$ square units." ], "src_path": [ "train/geometry/688.json", "train/geometry/688.json", "train/geometry/688.json", "train/geometry/688.json", "train/geometry/688.json", "train/geometry/688.json" ], "input": [ "The nine points of this grid are equally spaced horizontally and vertically. The distance between two neighboring points is 1 unit. What is the area, in square units, of the region where the two triangles overlap?\n\n[asy]\nsize(80);\ndot((0,0)); dot((0,1));dot((0,2));dot((1,0));dot((1,1));dot((1,2));dot((2,0));dot((2,1));dot((2,2));\ndraw((0,0)--(2,1)--(1,2)--cycle, linewidth(0.6));\ndraw((2,2)--(0,1)--(1,0)--cycle, linewidth(0.6));\n[/asy]", "The nine points of this grid are equally spaced horizontally and vertically. The distance between two neighboring points is 1 unit. What is the area, in square units, of the region where the two triangles overlap?\n\n[asy]\nsize(80);\ndot((0,0)); dot((0,1));dot((0,2));dot((1,0));dot((1,1));dot((1,2));dot((2,0));dot((2,1));dot((2,2));\ndraw((0,0)--(2,1)--(1,2)--cycle, linewidth(0.6));\ndraw((2,2)--(0,1)--(1,0)--cycle, linewidth(0.6));\n[/asy]", "The nine points of this grid are equally spaced horizontally and vertically. The distance between two neighboring points is 1 unit. What is the area, in square units, of the region where the two triangles overlap?\n\n[asy]\nsize(80);\ndot((0,0)); dot((0,1));dot((0,2));dot((1,0));dot((1,1));dot((1,2));dot((2,0));dot((2,1));dot((2,2));\ndraw((0,0)--(2,1)--(1,2)--cycle, linewidth(0.6));\ndraw((2,2)--(0,1)--(1,0)--cycle, linewidth(0.6));\n[/asy]", "The nine points of this grid are equally spaced horizontally and vertically. The distance between two neighboring points is 1 unit. What is the area, in square units, of the region where the two triangles overlap?\n\n[asy]\nsize(80);\ndot((0,0)); dot((0,1));dot((0,2));dot((1,0));dot((1,1));dot((1,2));dot((2,0));dot((2,1));dot((2,2));\ndraw((0,0)--(2,1)--(1,2)--cycle, linewidth(0.6));\ndraw((2,2)--(0,1)--(1,0)--cycle, linewidth(0.6));\n[/asy]", "The nine points of this grid are equally spaced horizontally and vertically. The distance between two neighboring points is 1 unit. What is the area, in square units, of the region where the two triangles overlap?\n\n[asy]\nsize(80);\ndot((0,0)); dot((0,1));dot((0,2));dot((1,0));dot((1,1));dot((1,2));dot((2,0));dot((2,1));dot((2,2));\ndraw((0,0)--(2,1)--(1,2)--cycle, linewidth(0.6));\ndraw((2,2)--(0,1)--(1,0)--cycle, linewidth(0.6));\n[/asy]", "The nine points of this grid are equally spaced horizontally and vertically. The distance between two neighboring points is 1 unit. What is the area, in square units, of the region where the two triangles overlap?\n\n[asy]\nsize(80);\ndot((0,0)); dot((0,1));dot((0,2));dot((1,0));dot((1,1));dot((1,2));dot((2,0));dot((2,1));dot((2,2));\ndraw((0,0)--(2,1)--(1,2)--cycle, linewidth(0.6));\ndraw((2,2)--(0,1)--(1,0)--cycle, linewidth(0.6));\n[/asy]" ], "output": [ "We color one of the triangles blue, and draw three blue segments connecting its points of intersection with the other triangle. [asy]\nsize(80);\ndot((0,0)); dot((0,1));dot((0,2));dot((1,0));dot((1,1));dot((1,2));dot((2,0));dot((2,1));dot((2,2));\ndraw((0,0)--(2,1)--(1,2)--cycle, blue+linewidth(0.6));\ndraw((2,2)--(0,1)--(1,0)--cycle, linewidth(0.6));\ndraw((.666,.333)--(1.333,1.666), blue+linewidth(0.6));\ndraw((.333,.666)--(1.666,1.333), blue+linewidth(0.6));\ndraw((1.333,.666)--(.666,1.333), blue+linewidth(0.6));\n[/asy] Because of the symmetry inherent in the grid and the two triangles (which are both isosceles), these three blue segments divide the blue triangle into congruent smaller triangles. The blue triangle contains 9 of these congruent smaller triangles.\n\nThe region of overlap of the two triangles is a hexagonal region. As per the diagram above, this hexagonal region contains 6 of these congruent smaller triangles. Thus, the area of the hexagonal region is $6/9=2/3$ of the area of one of the isosceles triangles. We compute the area of one isosceles triangle as follows:\n\n[asy]\nsize(100);\ndraw((0,0)--(2,0)--(2,2)--(0,2)--cycle);\ndraw((0,0)--(2,1)--(1,2)--cycle, linewidth(0.6));\nlabel(\"$A$\",2(0,0),SW); label(\"$B$\",2(1,0),SE); label(\"$C$\",2(1,1),NE); label(\"$D$\",2(0,1),NW); label(\"$E$\",2(.5,1),N); label(\"$F$\",2(1,.5),E);\n\n[/asy] Label points $A,B,C,D,E,F$ as above. To compute the area of this triangle ($\\triangle AEF$), notice how it is equal to the area of square $ABCD$ minus the areas of triangles $\\triangle ADE$, $\\triangle ABF$, and $\\triangle ECF$. The square has side length 2 units, so the area of $\\triangle ADE$ and $\\triangle ABF$ is $\\frac{1}{2}(2)(1) = 1$ and the area of $\\triangle ECF$ is $\\frac{1}{2}(1)(1)=\\frac{1}{2}$. The area of square $ABCD$ is $2^2=4$, so the area of triangle $\\triangle AEF$ is equal to $4 - 2(1) - \\frac{1}{2} = \\frac{3}{2}$.\n\nFinally, remember that the hexagonal region has area $2/3$ of the area of the triangle, or $\\frac{2}{3} \\cdot \\frac{3}{2} = 1$. Thus, the answer is $\\boxed{1}$.", "We color one of the triangles blue, and draw three blue segments connecting its points of intersection with the other triangle. [asy]\nsize(80);\ndot((0,0)); dot((0,1));dot((0,2));dot((1,0));dot((1,1));dot((1,2));dot((2,0));dot((2,1));dot((2,2));\ndraw((0,0)--(2,1)--(1,2)--cycle, blue+linewidth(0.6));\ndraw((2,2)--(0,1)--(1,0)--cycle, linewidth(0.6));\ndraw((.666,.333)--(1.333,1.666), blue+linewidth(0.6));\ndraw((.333,.666)--(1.666,1.333), blue+linewidth(0.6));\ndraw((1.333,.666)--(.666,1.333), blue+linewidth(0.6));\n[/asy] Because of the symmetry inherent in the grid and the two triangles (which are both isosceles), these three blue segments divide the blue triangle into congruent smaller triangles. The blue triangle contains 9 of these congruent smaller triangles.\n\nThe region of overlap of the two triangles is a hexagonal region. As per the diagram above, this hexagonal region contains 6 of these congruent smaller triangles. Thus, the area of the hexagonal region is $6/9=2/3$ of the area of one of the isosceles triangles. We compute the area of one isosceles triangle as follows:\n\n[asy]\nsize(100);\ndraw((0,0)--(2,0)--(2,2)--(0,2)--cycle);\ndraw((0,0)--(2,1)--(1,2)--cycle, linewidth(0.6));\nlabel(\"$A$\",2(0,0),SW); label(\"$B$\",2(1,0),SE); label(\"$C$\",2(1,1),NE); label(\"$D$\",2(0,1),NW); label(\"$E$\",2(.5,1),N); label(\"$F$\",2(1,.5),E);\n\n[/asy] Label points $A,B,C,D,E,F$ as above. To compute the area of this triangle ($\\triangle AEF$), notice how it is equal to the area of square $ABCD$ minus the areas of triangles $\\triangle ADE$, $\\triangle ABF$, and $\\triangle ECF$. The square has side length 2 units, so the area of $\\triangle ADE$ and $\\triangle ABF$ is $\\frac{1}{2}(2)(1) = 1$ and the area of $\\triangle ECF$ is $\\frac{1}{2}(1)(1)=\\frac{1}{2}$. The area of square $ABCD$ is $2^2=4$, so the area of triangle $\\triangle AEF$ is equal to $4 - 2(1) - \\frac{1}{2} = \\frac{3}{2}$.\n\nFinally, remember that the hexagonal region has area $2/3$ of the area of the triangle, or $\\frac{2}{3} \\cdot \\frac{3}{2} = 1$. Thus, the answer is $\\boxed{1}$.", "We color one of the triangles blue, and draw three blue segments connecting its points of intersection with the other triangle. [asy]\nsize(80);\ndot((0,0)); dot((0,1));dot((0,2));dot((1,0));dot((1,1));dot((1,2));dot((2,0));dot((2,1));dot((2,2));\ndraw((0,0)--(2,1)--(1,2)--cycle, blue+linewidth(0.6));\ndraw((2,2)--(0,1)--(1,0)--cycle, linewidth(0.6));\ndraw((.666,.333)--(1.333,1.666), blue+linewidth(0.6));\ndraw((.333,.666)--(1.666,1.333), blue+linewidth(0.6));\ndraw((1.333,.666)--(.666,1.333), blue+linewidth(0.6));\n[/asy] Because of the symmetry inherent in the grid and the two triangles (which are both isosceles), these three blue segments divide the blue triangle into congruent smaller triangles. The blue triangle contains 9 of these congruent smaller triangles.\n\nThe region of overlap of the two triangles is a hexagonal region. As per the diagram above, this hexagonal region contains 6 of these congruent smaller triangles. Thus, the area of the hexagonal region is $6/9=2/3$ of the area of one of the isosceles triangles. We compute the area of one isosceles triangle as follows:\n\n[asy]\nsize(100);\ndraw((0,0)--(2,0)--(2,2)--(0,2)--cycle);\ndraw((0,0)--(2,1)--(1,2)--cycle, linewidth(0.6));\nlabel(\"$A$\",2(0,0),SW); label(\"$B$\",2(1,0),SE); label(\"$C$\",2(1,1),NE); label(\"$D$\",2(0,1),NW); label(\"$E$\",2(.5,1),N); label(\"$F$\",2(1,.5),E);\n\n[/asy] Label points $A,B,C,D,E,F$ as above. To compute the area of this triangle ($\\triangle AEF$), notice how it is equal to the area of square $ABCD$ minus the areas of triangles $\\triangle ADE$, $\\triangle ABF$, and $\\triangle ECF$. The square has side length 2 units, so the area of $\\triangle ADE$ and $\\triangle ABF$ is $\\frac{1}{2}(2)(1) = 1$ and the area of $\\triangle ECF$ is $\\frac{1}{2}(1)(1)=\\frac{1}{2}$. The area of square $ABCD$ is $2^2=4$, so the area of triangle $\\triangle AEF$ is equal to $4 - 2(1) - \\frac{1}{2} = \\frac{3}{2}$.\n\nFinally, remember that the hexagonal region has area $2/3$ of the area of the triangle, or $\\frac{2}{3} \\cdot \\frac{3}{2} = 1$. Thus, the answer is $\\boxed{1}$.", "We color one of the triangles blue, and draw three blue segments connecting its points of intersection with the other triangle. [asy]\nsize(80);\ndot((0,0)); dot((0,1));dot((0,2));dot((1,0));dot((1,1));dot((1,2));dot((2,0));dot((2,1));dot((2,2));\ndraw((0,0)--(2,1)--(1,2)--cycle, blue+linewidth(0.6));\ndraw((2,2)--(0,1)--(1,0)--cycle, linewidth(0.6));\ndraw((.666,.333)--(1.333,1.666), blue+linewidth(0.6));\ndraw((.333,.666)--(1.666,1.333), blue+linewidth(0.6));\ndraw((1.333,.666)--(.666,1.333), blue+linewidth(0.6));\n[/asy] Because of the symmetry inherent in the grid and the two triangles (which are both isosceles), these three blue segments divide the blue triangle into congruent smaller triangles. The blue triangle contains 9 of these congruent smaller triangles.\n\nThe region of overlap of the two triangles is a hexagonal region. As per the diagram above, this hexagonal region contains 6 of these congruent smaller triangles. Thus, the area of the hexagonal region is $6/9=2/3$ of the area of one of the isosceles triangles. We compute the area of one isosceles triangle as follows:\n\n[asy]\nsize(100);\ndraw((0,0)--(2,0)--(2,2)--(0,2)--cycle);\ndraw((0,0)--(2,1)--(1,2)--cycle, linewidth(0.6));\nlabel(\"$A$\",2(0,0),SW); label(\"$B$\",2(1,0),SE); label(\"$C$\",2(1,1),NE); label(\"$D$\",2(0,1),NW); label(\"$E$\",2(.5,1),N); label(\"$F$\",2(1,.5),E);\n\n[/asy] Label points $A,B,C,D,E,F$ as above. To compute the area of this triangle ($\\triangle AEF$), notice how it is equal to the area of square $ABCD$ minus the areas of triangles $\\triangle ADE$, $\\triangle ABF$, and $\\triangle ECF$. The square has side length 2 units, so the area of $\\triangle ADE$ and $\\triangle ABF$ is $\\frac{1}{2}(2)(1) = 1$ and the area of $\\triangle ECF$ is $\\frac{1}{2}(1)(1)=\\frac{1}{2}$. The area of square $ABCD$ is $2^2=4$, so the area of triangle $\\triangle AEF$ is equal to $4 - 2(1) - \\frac{1}{2} = \\frac{3}{2}$.\n\nFinally, remember that the hexagonal region has area $2/3$ of the area of the triangle, or $\\frac{2}{3} \\cdot \\frac{3}{2} = 1$. Thus, the answer is $\\boxed{1}$.", "We color one of the triangles blue, and draw three blue segments connecting its points of intersection with the other triangle. [asy]\nsize(80);\ndot((0,0)); dot((0,1));dot((0,2));dot((1,0));dot((1,1));dot((1,2));dot((2,0));dot((2,1));dot((2,2));\ndraw((0,0)--(2,1)--(1,2)--cycle, blue+linewidth(0.6));\ndraw((2,2)--(0,1)--(1,0)--cycle, linewidth(0.6));\ndraw((.666,.333)--(1.333,1.666), blue+linewidth(0.6));\ndraw((.333,.666)--(1.666,1.333), blue+linewidth(0.6));\ndraw((1.333,.666)--(.666,1.333), blue+linewidth(0.6));\n[/asy] Because of the symmetry inherent in the grid and the two triangles (which are both isosceles), these three blue segments divide the blue triangle into congruent smaller triangles. The blue triangle contains 9 of these congruent smaller triangles.\n\nThe region of overlap of the two triangles is a hexagonal region. As per the diagram above, this hexagonal region contains 6 of these congruent smaller triangles. Thus, the area of the hexagonal region is $6/9=2/3$ of the area of one of the isosceles triangles. We compute the area of one isosceles triangle as follows:\n\n[asy]\nsize(100);\ndraw((0,0)--(2,0)--(2,2)--(0,2)--cycle);\ndraw((0,0)--(2,1)--(1,2)--cycle, linewidth(0.6));\nlabel(\"$A$\",2(0,0),SW); label(\"$B$\",2(1,0),SE); label(\"$C$\",2(1,1),NE); label(\"$D$\",2(0,1),NW); label(\"$E$\",2(.5,1),N); label(\"$F$\",2(1,.5),E);\n\n[/asy] Label points $A,B,C,D,E,F$ as above. To compute the area of this triangle ($\\triangle AEF$), notice how it is equal to the area of square $ABCD$ minus the areas of triangles $\\triangle ADE$, $\\triangle ABF$, and $\\triangle ECF$. The square has side length 2 units, so the area of $\\triangle ADE$ and $\\triangle ABF$ is $\\frac{1}{2}(2)(1) = 1$ and the area of $\\triangle ECF$ is $\\frac{1}{2}(1)(1)=\\frac{1}{2}$. The area of square $ABCD$ is $2^2=4$, so the area of triangle $\\triangle AEF$ is equal to $4 - 2(1) - \\frac{1}{2} = \\frac{3}{2}$.\n\nFinally, remember that the hexagonal region has area $2/3$ of the area of the triangle, or $\\frac{2}{3} \\cdot \\frac{3}{2} = 1$. Thus, the answer is $\\boxed{1}$.", "We color one of the triangles blue, and draw three blue segments connecting its points of intersection with the other triangle. [asy]\nsize(80);\ndot((0,0)); dot((0,1));dot((0,2));dot((1,0));dot((1,1));dot((1,2));dot((2,0));dot((2,1));dot((2,2));\ndraw((0,0)--(2,1)--(1,2)--cycle, blue+linewidth(0.6));\ndraw((2,2)--(0,1)--(1,0)--cycle, linewidth(0.6));\ndraw((.666,.333)--(1.333,1.666), blue+linewidth(0.6));\ndraw((.333,.666)--(1.666,1.333), blue+linewidth(0.6));\ndraw((1.333,.666)--(.666,1.333), blue+linewidth(0.6));\n[/asy] Because of the symmetry inherent in the grid and the two triangles (which are both isosceles), these three blue segments divide the blue triangle into congruent smaller triangles. The blue triangle contains 9 of these congruent smaller triangles.\n\nThe region of overlap of the two triangles is a hexagonal region. As per the diagram above, this hexagonal region contains 6 of these congruent smaller triangles. Thus, the area of the hexagonal region is $6/9=2/3$ of the area of one of the isosceles triangles. We compute the area of one isosceles triangle as follows:\n\n[asy]\nsize(100);\ndraw((0,0)--(2,0)--(2,2)--(0,2)--cycle);\ndraw((0,0)--(2,1)--(1,2)--cycle, linewidth(0.6));\nlabel(\"$A$\",2(0,0),SW); label(\"$B$\",2(1,0),SE); label(\"$C$\",2(1,1),NE); label(\"$D$\",2(0,1),NW); label(\"$E$\",2(.5,1),N); label(\"$F$\",2(1,.5),E);\n\n[/asy] Label points $A,B,C,D,E,F$ as above. To compute the area of this triangle ($\\triangle AEF$), notice how it is equal to the area of square $ABCD$ minus the areas of triangles $\\triangle ADE$, $\\triangle ABF$, and $\\triangle ECF$. The square has side length 2 units, so the area of $\\triangle ADE$ and $\\triangle ABF$ is $\\frac{1}{2}(2)(1) = 1$ and the area of $\\triangle ECF$ is $\\frac{1}{2}(1)(1)=\\frac{1}{2}$. The area of square $ABCD$ is $2^2=4$, so the area of triangle $\\triangle AEF$ is equal to $4 - 2(1) - \\frac{1}{2} = \\frac{3}{2}$.\n\nFinally, remember that the hexagonal region has area $2/3$ of the area of the triangle, or $\\frac{2}{3} \\cdot \\frac{3}{2} = 1$. Thus, the answer is $\\boxed{1}$." ], "rewards": [ 0.0, 0.0, 0.0, 0.0, 0.0, 0.0 ] }