{
  "index": "1990-A-3",
  "type": "GEO",
  "tag": [
    "GEO",
    "NT",
    "COMB"
  ],
  "difficulty": "",
  "question": "which are collinear) have integer coordinates must have area greater than\nor equal to 5/2.",
  "solution": "Solution. A lattice polygon is a plane polygon whose vertices are lattice points, i.e., points with integer coordinates. The area of any convex lattice polygon has area equal to half an integer: this follows from Pick's Theorem mentioned in the second remark below; alternatively, by subdivision one can reduce to the case of a triangle, in which case the statement follows from the first remark below.\n\nConsider a convex lattice pentagon \\( A B C D E \\) of minimal area. Since the area is always half an integer, the minimum exists. If the interior of side \\( A B \\) contains a lattice point \\( F \\), then \\( A F C D E \\) is a convex lattice pentagon with smaller area, contradicting the choice of \\( A B C D E \\). (As is standard, vertices are listed in order around each polygon.) Applying this argument to each side, we may assume that all boundary lattice points are vertices.\n\nSeparate the vertices into four classes according to the parity of their coordinates. By the Pigeonhole Principle, one class must contain at least two vertices. The midpoint \\( M \\) between two such vertices has integer coordinates. By the previous paragraph, these two vertices cannot form a side of the polygon. Also, the pentagon is convex, so \\( M \\) is in the interior of the pentagon. Connecting \\( M \\) to the vertices divides the polygon into 5 triangles, each of area at least \\( 1 / 2 \\), so the whole polygon has area at least \\( 5 / 2 \\).\n\nRemark. The bound \\( 5 / 2 \\) cannot be improved: the polygon with vertices \\( (0,0) \\), \\( (1,0),(2,1),(1,2),(0,1) \\) is convex and has area \\( 5 / 2 \\) (see Figure 14).\n\nRemark. The area of a plane triangle with vertices \\( \\left(x_{1}, y_{1}\\right),\\left(x_{2}, y_{2}\\right),\\left(x_{3}, y_{3}\\right) \\) equals\n\\[\n\\frac{1}{2}\\left|\\operatorname{det}\\left(\\begin{array}{lll}\nx_{1} & y_{1} & 1 \\\\\nx_{2} & y_{2} & 1 \\\\\nx_{3} & y_{3} & 1\n\\end{array}\\right)\\right|\n\\]\n\nIn particular, if \\( x_{i}, y_{i} \\in \\mathbb{Z} \\), the area is half an integer.\nFor a formula for the volume of an \\( n \\)-dimensional simplex in \\( \\mathbb{R}^{n} \\), see Solution 5 to 1993B5.\n\nRemark (Pick's Theorem). Given a lattice polygon, let \\( i \\) be the number of internal lattice points and let \\( b \\) be the number of boundary lattice points. Pick's Theorem [Lar1, p. 68] states that the area of the polygon equals \\( i+b / 2-1 \\).\n\nIn the previous solution, we could have used Pick's Theorem in two places: in the first paragraph to prove that a lattice polygon has half-integer area (even if it is not convex), and as a substitute for the last sentence, using \\( i \\geq 1 \\) and \\( b=5 \\).\n\nRelated question. Problem 1981A6 [PutnamII, p. 37] is similar:\nSuppose that each of the vertices of \\( \\triangle A B C \\) is a lattice point in the \\( (x, y) \\)-plane and that there is exactly one lattice point \\( P \\) in the interior of the triangle.\nThe line \\( A P \\) is extended to meet \\( B C \\) at \\( E \\). Determine the largest possible value for the ratio of the lengths of segments \\( |A P| /|P E| \\).\nHere is a solution to 1981A6 using Pick's Theorem. (See [PutnamII, p. 118] for a non-Pick solution.)\n\nWe may reduce to the case where \\( B C \\) has no lattice points on it other than \\( B, C \\), and possibly \\( E \\), by replacing the base \\( B C \\) with the shortest segment along it with lattice endpoints and containing \\( E \\) in its interior.\n\nCase 1: \\( E \\) is a lattice point. Then the reflection of \\( E \\) across \\( P \\) is also a lattice point, so it must coincide with \\( A \\), and \\( |A P| /|P E|=1 \\).\n\nCase 2: \\( E \\) is not a lattice point. Without loss of generality (by applying an affine transformation preserving the lattice), we may assume \\( B=(0,0) \\) and \\( C=(1,0) \\). If \\( A=(x, y), y>0 \\), then \\( x \\neq 0,1 \\), and by Pick's Theorem,\n\\[\n\\begin{aligned}\ny & =2+\\#\\{\\text { boundary lattice points }\\}-2 \\\\\n& =d+e+1\n\\end{aligned}\n\\]\nwhere \\( d=\\operatorname{gcd}(x, y), e=\\operatorname{gcd}(1-x, y) \\). Since \\( d \\) and \\( e \\) are relatively prime, de divides \\( y \\), so \\( d e \\leq d+e+1 \\), or equivalently \\( (d-1)(e-1) \\leq 2 \\). We have several subcases:\n- If \\( d=1 \\), then \\( y=2+e \\) is divisible by \\( e \\), so \\( e=1 \\) or 2 . If \\( e=1 \\), then \\( y=2 \\), and the ratio \\( |A P| /|P E| \\) is 2 . If \\( e=2 \\), then \\( y=3 \\), and the ratio is 3 .\n- If \\( e=1 \\), then essentially the same argument gives a ratio of 2 or 3 .\n- The case \\( d=e=2 \\) is not allowed, since \\( d \\) and \\( e \\) are coprime.\n- If \\( d=3 \\) and \\( e=2 \\), then \\( y=6 \\), giving a ratio of 5 .\n- If \\( d=2 \\) and \\( e=3 \\), then the ratio is also 5 .\n\nHence the maximum is 5 . This argument can be refined to show that equality is achieved only for one triangle (up to automorphisms of the plane preserving lattice points).\n\nRemark. The group of linear transformations of the plane preserving the lattice points and fixing the origin is the group \\( \\mathrm{GL}_{2}(\\mathbb{Z}) \\) defined in the introduction. It is important in number theory and related fields.\n\nRemark. Fix integers \\( d \\geq 2 \\) and \\( k \\geq 1 \\). It follows easily from [He] that up to the action of \\( \\mathrm{GL}_{d}(\\mathbb{Z}) \\) and translation by lattice points, there are only finitely many convex lattice polytopes in \\( \\mathbb{R}^{n} \\) having exactly \\( k \\) interior lattice points. For \\( d=2 \\) and \\( k=1 \\), we get 16 polygons; see [PRV] for a figure showing all of them. Five of these 16 are triangles, and checking each case gives another proof that \\( |A P| /|P E| \\leq 5 \\), and that equality is possible for only one of the five triangles.",
  "vars": [
    "A",
    "B",
    "C",
    "D",
    "E",
    "F",
    "M",
    "P",
    "x",
    "x_1",
    "x_2",
    "x_3",
    "y",
    "y_1",
    "y_2",
    "y_3",
    "i",
    "b"
  ],
  "params": [
    "d",
    "e",
    "n",
    "k",
    "R"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "A": "pointalpha",
        "B": "pointbeta",
        "C": "pointgamma",
        "D": "pointdelta",
        "E": "pointepsilon",
        "F": "pointzeta",
        "M": "midpoint",
        "P": "interiorp",
        "x": "coordx",
        "x_1": "coordxone",
        "x_2": "coordxtwo",
        "x_3": "coordxthree",
        "y": "coordy",
        "y_1": "coordyone",
        "y_2": "coordytwo",
        "y_3": "coordythree",
        "i": "interiorcount",
        "b": "boundarycount",
        "d": "gcdfirst",
        "e": "gcdsecond",
        "n": "dimensionn",
        "k": "interiork",
        "R": "regionr"
      },
      "question": "which are collinear) have integer coordinates must have area greater than\nor equal to 5/2.",
      "solution": "Solution. A lattice polygon is a plane polygon whose vertices are lattice points, i.e., points with integer coordinates. The area of any convex lattice polygon has area equal to half an integer: this follows from Pick's Theorem mentioned in the second remark below; alternatively, by subdivision one can reduce to the case of a triangle, in which case the statement follows from the first remark below.\n\nConsider a convex lattice pentagon \\( pointalpha\\ pointbeta\\ pointgamma\\ pointdelta\\ pointepsilon \\) of minimal area. Since the area is always half an integer, the minimum exists. If the interior of side \\( pointalpha\\ pointbeta \\) contains a lattice point \\( pointzeta \\), then \\( pointalpha\\ pointzeta\\ pointgamma\\ pointdelta\\ pointepsilon \\) is a convex lattice pentagon with smaller area, contradicting the choice of \\( pointalpha\\ pointbeta\\ pointgamma\\ pointdelta\\ pointepsilon \\). (As is standard, vertices are listed in order around each polygon.) Applying this argument to each side, we may assume that all boundary lattice points are vertices.\n\nSeparate the vertices into four classes according to the parity of their coordinates. By the Pigeonhole Principle, one class must contain at least two vertices. The midpoint \\( midpoint \\) between two such vertices has integer coordinates. By the previous paragraph, these two vertices cannot form a side of the polygon. Also, the pentagon is convex, so \\( midpoint \\) is in the interior of the pentagon. Connecting \\( midpoint \\) to the vertices divides the polygon into 5 triangles, each of area at least \\( 1 / 2 \\), so the whole polygon has area at least \\( 5 / 2 \\).\n\nRemark. The bound \\( 5 / 2 \\) cannot be improved: the polygon with vertices \\( (0,0) \\), \\( (1,0),(2,1),(1,2),(0,1) \\) is convex and has area \\( 5 / 2 \\) (see Figure 14).\n\nRemark. The area of a plane triangle with vertices \\( \\left(coordxone, coordyone\\right),\\left(coordxtwo, coordytwo\\right),\\left(coordxthree, coordythree\\right) \\) equals\n\\[\n\\frac{1}{2}\\left|\\operatorname{det}\\left(\\begin{array}{lll}\ncoordxone & coordyone & 1 \\\\\ncoordxtwo & coordytwo & 1 \\\\\ncoordxthree & coordythree & 1\n\\end{array}\\right)\\right|\n\\]\n\nIn particular, if \\( coordx_{interiorcount}, coordy_{interiorcount} \\in \\mathbb{Z} \\), the area is half an integer.\nFor a formula for the volume of an \\( dimensionn \\)-dimensional simplex in \\( \\mathbb{R}^{dimensionn} \\), see Solution 5 to 1993B5.\n\nRemark (Pick's Theorem). Given a lattice polygon, let \\( interiorcount \\) be the number of internal lattice points and let \\( boundarycount \\) be the number of boundary lattice points. Pick's Theorem [Lar1, p. 68] states that the area of the polygon equals \\( interiorcount+boundarycount / 2-1 \\).\n\nIn the previous solution, we could have used Pick's Theorem in two places: in the first paragraph to prove that a lattice polygon has half-integer area (even if it is not convex), and as a substitute for the last sentence, using \\( interiorcount \\geq 1 \\) and \\( boundarycount=5 \\).\n\nRelated question. Problem 1981A6 [PutnamII, p. 37] is similar:\nSuppose that each of the vertices of \\( \\triangle pointalpha\\ pointbeta\\ pointgamma \\) is a lattice point in the \\( (coordx, coordy) \\)-plane and that there is exactly one lattice point \\( interiorp \\) in the interior of the triangle.\nThe line \\( pointalpha\\ interiorp \\) is extended to meet \\( pointbeta\\ pointgamma \\) at \\( pointepsilon \\). Determine the largest possible value for the ratio of the lengths of segments \\( |pointalpha\\ interiorp| /|interiorp\\ pointepsilon| \\).\nHere is a solution to 1981A6 using Pick's Theorem. (See [PutnamII, p. 118] for a non-Pick solution.)\n\nWe may reduce to the case where \\( pointbeta\\ pointgamma \\) has no lattice points on it other than \\( pointbeta, pointgamma \\), and possibly \\( pointepsilon \\), by replacing the base \\( pointbeta\\ pointgamma \\) with the shortest segment along it with lattice endpoints and containing \\( pointepsilon \\) in its interior.\n\nCase 1: \\( pointepsilon \\) is a lattice point. Then the reflection of \\( pointepsilon \\) across \\( interiorp \\) is also a lattice point, so it must coincide with \\( pointalpha \\), and \\( |pointalpha\\ interiorp| /|interiorp\\ pointepsilon|=1 \\).\n\nCase 2: \\( pointepsilon \\) is not a lattice point. Without loss of generality (by applying an affine transformation preserving the lattice), we may assume \\( pointbeta=(0,0) \\) and \\( pointgamma=(1,0) \\). If \\( pointalpha=(coordx, coordy), coordy>0 \\), then \\( coordx \\neq 0,1 \\), and by Pick's Theorem,\n\\[\n\\begin{aligned}\ncoordy & =2+\\#\\{\\text { boundary lattice points }\\}-2 \\\\\n& =gcdfirst+gcdsecond+1\n\\end{aligned}\n\\]\nwhere \\( gcdfirst=\\operatorname{gcd}(coordx, coordy), gcdsecond=\\operatorname{gcd}(1-coordx, coordy) \\). Since \\( gcdfirst \\) and \\( gcdsecond \\) are relatively prime, gcdfirst gcdsecond divides \\( coordy \\), so \\( gcdfirst gcdsecond \\leq gcdfirst+gcdsecond+1 \\), or equivalently \\( (gcdfirst-1)(gcdsecond-1) \\leq 2 \\). We have several subcases:\n- If \\( gcdfirst=1 \\), then \\( coordy=2+gcdsecond \\) is divisible by \\( gcdsecond \\), so \\( gcdsecond=1 \\) or 2 . If \\( gcdsecond=1 \\), then \\( coordy=2 \\), and the ratio \\( |pointalpha\\ interiorp| /|interiorp\\ pointepsilon| \\) is 2 . If \\( gcdsecond=2 \\), then \\( coordy=3 \\), and the ratio is 3 .\n- If \\( gcdsecond=1 \\), then essentially the same argument gives a ratio of 2 or 3 .\n- The case \\( gcdfirst=gcdsecond=2 \\) is not allowed, since \\( gcdfirst \\) and \\( gcdsecond \\) are coprime.\n- If \\( gcdfirst=3 \\) and \\( gcdsecond=2 \\), then \\( coordy=6 \\), giving a ratio of 5 .\n- If \\( gcdfirst=2 \\) and \\( gcdsecond=3 \\), then the ratio is also 5 .\n\nHence the maximum is 5 . This argument can be refined to show that equality is achieved only for one triangle (up to automorphisms of the plane preserving lattice points).\n\nRemark. The group of linear transformations of the plane preserving the lattice points and fixing the origin is the group \\( \\mathrm{GL}_{2}(\\mathbb{Z}) \\) defined in the introduction. It is important in number theory and related fields.\n\nRemark. Fix integers \\( gcdfirst \\geq 2 \\) and \\( interiork \\geq 1 \\). It follows easily from [He] that up to the action of \\( \\mathrm{GL}_{gcdfirst}(\\mathbb{Z}) \\) and translation by lattice points, there are only finitely many convex lattice polytopes in \\( \\mathbb{R}^{dimensionn} \\) having exactly \\( interiork \\) interior lattice points. For \\( gcdfirst=2 \\) and \\( interiork=1 \\), we get 16 polygons; see [PRV] for a figure showing all of them. Five of these 16 are triangles, and checking each case gives another proof that \\( |pointalpha\\ interiorp| /|interiorp\\ pointepsilon| \\leq 5 \\), and that equality is possible for only one of the five triangles."
    },
    "descriptive_long_confusing": {
      "map": {
        "A": "watermelon",
        "B": "toothbrush",
        "C": "sailplane",
        "D": "briefcase",
        "E": "spotlight",
        "F": "rainstorm",
        "M": "goldfish",
        "P": "bookshelf",
        "x": "lemonade",
        "x_1": "masquerade",
        "x_2": "hairbrush",
        "x_3": "snowflake",
        "y": "dishwasher",
        "y_1": "chameleon",
        "y_2": "toothpick",
        "y_3": "nightshade",
        "i": "marshmallow",
        "b": "microscope",
        "d": "skateboard",
        "e": "rhinoceros",
        "n": "escapement",
        "k": "thunderbolt",
        "R": "candlestick"
      },
      "question": "which are collinear) have integer coordinates must have area greater than\nor equal to 5/2.",
      "solution": "Solution. A lattice polygon is a plane polygon whose vertices are lattice points, i.e., points with integer coordinates. The area of any convex lattice polygon has area equal to half an integer: this follows from Pick's Theorem mentioned in the second remark below; alternatively, by subdivision one can reduce to the case of a triangle, in which case the statement follows from the first remark below.\n\nConsider a convex lattice pentagon \\( watermelon toothbrush sailplane briefcase spotlight \\) of minimal area. Since the area is always half an integer, the minimum exists. If the interior of side \\( watermelon \\, toothbrush \\) contains a lattice point \\( rainstorm \\), then \\( watermelon \\, rainstorm \\, sailplane \\, briefcase \\, spotlight \\) is a convex lattice pentagon with smaller area, contradicting the choice of \\( watermelon \\, toothbrush \\, sailplane \\, briefcase \\, spotlight \\). (As is standard, vertices are listed in order around each polygon.) Applying this argument to each side, we may assume that all boundary lattice points are vertices.\n\nSeparate the vertices into four classes according to the parity of their coordinates. By the Pigeonhole Principle, one class must contain at least two vertices. The midpoint \\( goldfish \\) between two such vertices has integer coordinates. By the previous paragraph, these two vertices cannot form a side of the polygon. Also, the pentagon is convex, so \\( goldfish \\) is in the interior of the pentagon. Connecting \\( goldfish \\) to the vertices divides the polygon into 5 triangles, each of area at least \\( 1 / 2 \\), so the whole polygon has area at least \\( 5 / 2 \\).\n\nRemark. The bound \\( 5 / 2 \\) cannot be improved: the polygon with vertices \\( (0,0) \\), \\( (1,0),(2,1),(1,2),(0,1) \\) is convex and has area \\( 5 / 2 \\) (see Figure 14).\n\nRemark. The area of a plane triangle with vertices \\( \\left(masquerade, chameleon\\right),\\left(hairbrush, toothpick\\right),\\left(snowflake, nightshade\\right) \\) equals\n\\[\n\\frac{1}{2}\\left|\\operatorname{det}\\left(\\begin{array}{lll}\nmasquerade & chameleon & 1 \\\\\nhairbrush & toothpick & 1 \\\\\nsnowflake & nightshade & 1\n\\end{array}\\right)\\right|\n\\]\n\nIn particular, if \\( lemonade_{marshmallow}, dishwasher_{marshmallow} \\in \\mathbb{Z} \\), the area is half an integer.\nFor a formula for the volume of an \\( escapement \\)-dimensional simplex in \\( \\mathbb{R}^{escapement} \\), see Solution 5 to 1993B5.\n\nRemark (Pick's Theorem). Given a lattice polygon, let \\( marshmallow \\) be the number of internal lattice points and let \\( microscope \\) be the number of boundary lattice points. Pick's Theorem [Lar1, p. 68] states that the area of the polygon equals \\( marshmallow+microscope / 2-1 \\).\n\nIn the previous solution, we could have used Pick's Theorem in two places: in the first paragraph to prove that a lattice polygon has half-integer area (even if it is not convex), and as a substitute for the last sentence, using \\( marshmallow \\geq 1 \\) and \\( microscope=5 \\).\n\nRelated question. Problem 1981A6 [PutnamII, p. 37] is similar:\nSuppose that each of the vertices of \\( \\triangle watermelon toothbrush sailplane \\) is a lattice point in the \\( (lemonade, dishwasher) \\)-plane and that there is exactly one lattice point \\( bookshelf \\) in the interior of the triangle.\nThe line \\( watermelon \\, bookshelf \\) is extended to meet \\( toothbrush sailplane \\) at \\( spotlight \\). Determine the largest possible value for the ratio of the lengths of segments \\( |watermelon \\, bookshelf| /|bookshelf \\, spotlight| \\).\nHere is a solution to 1981A6 using Pick's Theorem. (See [PutnamII, p. 118] for a non-Pick solution.)\n\nWe may reduce to the case where \\( toothbrush \\, sailplane \\) has no lattice points on it other than \\( toothbrush, sailplane \\), and possibly \\( spotlight \\), by replacing the base \\( toothbrush \\, sailplane \\) with the shortest segment along it with lattice endpoints and containing \\( spotlight \\) in its interior.\n\nCase 1: \\( spotlight \\) is a lattice point. Then the reflection of \\( spotlight \\) across \\( bookshelf \\) is also a lattice point, so it must coincide with \\( watermelon \\), and \\( |watermelon \\, bookshelf| /|bookshelf \\, spotlight|=1 \\).\n\nCase 2: \\( spotlight \\) is not a lattice point. Without loss of generality (by applying an affine transformation preserving the lattice), we may assume \\( toothbrush=(0,0) \\) and \\( sailplane=(1,0) \\). If \\( watermelon=(lemonade, dishwasher), dishwasher>0 \\), then \\( lemonade \\neq 0,1 \\), and by Pick's Theorem,\n\\[\n\\begin{aligned}\ndishwasher & =2+\\#\\{\\text { boundary lattice points }\\}-2 \\\\\n& =skateboard+rhinoceros+1\n\\end{aligned}\n\\]\nwhere \\( skateboard=\\operatorname{gcd}(lemonade, dishwasher), rhinoceros=\\operatorname{gcd}(1-lemonade, dishwasher) \\). Since \\( skateboard \\) and \\( rhinoceros \\) are relatively prime, skateboard\\,rhinoceros divides \\( dishwasher \\), so \\( skateboard \\, rhinoceros \\leq skateboard+rhinoceros+1 \\), or equivalently \\( (skateboard-1)(rhinoceros-1) \\leq 2 \\). We have several subcases:\n- If \\( skateboard=1 \\), then \\( dishwasher=2+rhinoceros \\) is divisible by \\( rhinoceros \\), so \\( rhinoceros=1 \\) or 2. If \\( rhinoceros=1 \\), then \\( dishwasher=2 \\), and the ratio \\( |watermelon \\, bookshelf| /|bookshelf \\, spotlight| \\) is 2. If \\( rhinoceros=2 \\), then \\( dishwasher=3 \\), and the ratio is 3.\n- If \\( rhinoceros=1 \\), then essentially the same argument gives a ratio of 2 or 3.\n- The case \\( skateboard=rhinoceros=2 \\) is not allowed, since \\( skateboard \\) and \\( rhinoceros \\) are coprime.\n- If \\( skateboard=3 \\) and \\( rhinoceros=2 \\), then \\( dishwasher=6 \\), giving a ratio of 5.\n- If \\( skateboard=2 \\) and \\( rhinoceros=3 \\), then the ratio is also 5.\n\nHence the maximum is 5. This argument can be refined to show that equality is achieved only for one triangle (up to automorphisms of the plane preserving lattice points).\n\nRemark. The group of linear transformations of the plane preserving the lattice points and fixing the origin is the group \\( \\mathrm{GL}_{2}(\\mathbb{Z}) \\) defined in the introduction. It is important in number theory and related fields.\n\nRemark. Fix integers \\( skateboard \\geq 2 \\) and \\( thunderbolt \\geq 1 \\). It follows easily from [He] that up to the action of \\( \\mathrm{GL}_{skateboard}(\\mathbb{Z}) \\) and translation by lattice points, there are only finitely many convex lattice polytopes in \\( \\mathbb{R}^{escapement} \\) having exactly \\( thunderbolt \\) interior lattice points. For \\( skateboard=2 \\) and \\( thunderbolt=1 \\), we get 16 polygons; see [PRV] for a figure showing all of them. Five of these 16 are triangles, and checking each case gives another proof that \\( |watermelon \\, bookshelf| /|bookshelf \\, spotlight| \\leq 5 \\), and that equality is possible for only one of the five triangles."
    },
    "descriptive_long_misleading": {
      "map": {
        "A": "centerpoint",
        "B": "midpoint",
        "C": "interiorpoint",
        "D": "medialpoint",
        "E": "corepoint",
        "F": "emptypoint",
        "M": "edgepoint",
        "P": "boundarypoint",
        "x": "magnitude",
        "x_1": "shiftone",
        "x_2": "shifttwo",
        "x_3": "shiftthree",
        "y": "direction",
        "y_1": "angleone",
        "y_2": "angletwo",
        "y_3": "anglethree",
        "i": "boundarycount",
        "b": "interiorcount",
        "d": "lcmvalue",
        "e": "lcmother",
        "n": "singular",
        "k": "emptiness",
        "R": "imaginary"
      },
      "question": "which are collinear) have integer coordinates must have area greater than or equal to 5/2.",
      "solution": "Solution. A lattice polygon is a plane polygon whose vertices are lattice points, i.e., points with integer coordinates. The area of any convex lattice polygon has area equal to half an integer: this follows from Pick's Theorem mentioned in the second remark below; alternatively, by subdivision one can reduce to the case of a triangle, in which case the statement follows from the first remark below.\n\nConsider a convex lattice pentagon \\( centerpoint\\ midpoint\\ interiorpoint\\ medialpoint\\ corepoint \\) of minimal area. Since the area is always half an integer, the minimum exists. If the interior of side \\( centerpoint midpoint \\) contains a lattice point \\( emptypoint \\), then \\( centerpoint\\ emptypoint\\ interiorpoint\\ medialpoint\\ corepoint \\) is a convex lattice pentagon with smaller area, contradicting the choice of \\( centerpoint\\ midpoint\\ interiorpoint\\ medialpoint\\ corepoint \\). (As is standard, vertices are listed in order around each polygon.) Applying this argument to each side, we may assume that all boundary lattice points are vertices.\n\nSeparate the vertices into four classes according to the parity of their coordinates. By the Pigeonhole Principle, one class must contain at least two vertices. The midpoint \\( edgepoint \\) between two such vertices has integer coordinates. By the previous paragraph, these two vertices cannot form a side of the polygon. Also, the pentagon is convex, so \\( edgepoint \\) is in the interior of the pentagon. Connecting \\( edgepoint \\) to the vertices divides the polygon into 5 triangles, each of area at least \\( 1 / 2 \\), so the whole polygon has area at least \\( 5 / 2 \\).\n\nRemark. The bound \\( 5 / 2 \\) cannot be improved: the polygon with vertices \\( (0,0) \\), \\( (1,0),(2,1),(1,2),(0,1) \\) is convex and has area \\( 5 / 2 \\) (see Figure 14).\n\nRemark. The area of a plane triangle with vertices \\( \\left(shiftone, angleone\\right),\\left(shifttwo, angletwo\\right),\\left(shiftthree, anglethree\\right) \\) equals\n\\[\n\\frac{1}{2}\\left|\\operatorname{det}\\left(\\begin{array}{lll}\nshiftone & angleone & 1 \\\\\nshifttwo & angletwo & 1 \\\\\nshiftthree & anglethree & 1\n\\end{array}\\right)\\right|\n\\]\n\nIn particular, if \\( shiftone, angleone \\in \\mathbb{Z} \\), the area is half an integer.\nFor a formula for the volume of an \\( singular \\)-dimensional simplex in \\( \\mathbb{R}^{singular} \\), see Solution 5 to 1993B5.\n\nRemark (Pick's Theorem). Given a lattice polygon, let \\( boundarycount \\) be the number of internal lattice points and let \\( interiorcount \\) be the number of boundary lattice points. Pick's Theorem [Lar1, p. 68] states that the area of the polygon equals \\( boundarycount + interiorcount / 2 - 1 \\).\n\nIn the previous solution, we could have used Pick's Theorem in two places: in the first paragraph to prove that a lattice polygon has half-integer area (even if it is not convex), and as a substitute for the last sentence, using \\( boundarycount \\geq 1 \\) and \\( interiorcount =5 \\).\n\nRelated question. Problem 1981A6 [PutnamII, p. 37] is similar:\nSuppose that each of the vertices of \\( \\triangle centerpoint\\ midpoint\\ interiorpoint \\) is a lattice point in the \\( (x, y) \\)-plane and that there is exactly one lattice point \\( boundarypoint \\) in the interior of the triangle.\nThe line \\( centerpoint\\ boundarypoint \\) is extended to meet \\( midpoint\\ interiorpoint \\) at \\( corepoint \\). Determine the largest possible value for the ratio of the lengths of segments \\( |centerpoint\\ boundarypoint| /|boundarypoint\\ corepoint| \\).\nHere is a solution to 1981A6 using Pick's Theorem. (See [PutnamII, p. 118] for a non-Pick solution.)\n\nWe may reduce to the case where \\( midpoint\\ interiorpoint \\) has no lattice points on it other than \\( midpoint, interiorpoint \\), and possibly \\( corepoint \\), by replacing the base \\( midpoint\\ interiorpoint \\) with the shortest segment along it with lattice endpoints and containing \\( corepoint \\) in its interior.\n\nCase 1: \\( corepoint \\) is a lattice point. Then the reflection of \\( corepoint \\) across \\( boundarypoint \\) is also a lattice point, so it must coincide with \\( centerpoint \\), and \\( |centerpoint\\ boundarypoint| /|boundarypoint\\ corepoint|=1 \\).\n\nCase 2: \\( corepoint \\) is not a lattice point. Without loss of generality (by applying an affine transformation preserving the lattice), we may assume \\( midpoint=(0,0) \\) and \\( interiorpoint=(1,0) \\). If \\( centerpoint=(magnitude, direction), direction>0 \\), then \\( magnitude \\neq 0,1 \\), and by Pick's Theorem,\n\\[\n\\begin{aligned}\ndirection & =2+\\#\\{\\text { boundary lattice points }\\}-2 \\\\\n& =lcmvalue + lcmother + 1\n\\end{aligned}\n\\]\nwhere \\( lcmvalue=\\operatorname{gcd}(magnitude, direction),\\; lcmother=\\operatorname{gcd}(1-magnitude, direction) \\). Since \\( lcmvalue \\) and \\( lcmother \\) are relatively prime, \\( lcmvalue lcmother \\) divides \\( direction \\), so \\( lcmvalue lcmother \\leq lcmvalue + lcmother + 1 \\), or equivalently \\( (lcmvalue-1)(lcmother-1) \\leq 2 \\). We have several subcases:\n- If \\( lcmvalue =1 \\), then \\( direction =2+lcmother \\) is divisible by \\( lcmother \\), so \\( lcmother =1 \\) or \\( 2 \\). If \\( lcmother =1 \\), then \\( direction =2 \\), and the ratio \\( |centerpoint\\ boundarypoint| /|boundarypoint\\ corepoint| \\) is \\( 2 \\). If \\( lcmother =2 \\), then \\( direction =3 \\), and the ratio is \\( 3 \\).\n- If \\( lcmother =1 \\), then essentially the same argument gives a ratio of \\( 2 \\) or \\( 3 \\).\n- The case \\( lcmvalue = lcmother =2 \\) is not allowed, since \\( lcmvalue \\) and \\( lcmother \\) are coprime.\n- If \\( lcmvalue =3 \\) and \\( lcmother =2 \\), then \\( direction =6 \\), giving a ratio of \\( 5 \\).\n- If \\( lcmvalue =2 \\) and \\( lcmother =3 \\), then the ratio is also \\( 5 \\).\n\nHence the maximum is \\( 5 \\). This argument can be refined to show that equality is achieved only for one triangle (up to automorphisms of the plane preserving lattice points).\n\nRemark. The group of linear transformations of the plane preserving the lattice points and fixing the origin is the group \\( \\mathrm{GL}_{2}(\\mathbb{Z}) \\) defined in the introduction. It is important in number theory and related fields.\n\nRemark. Fix integers \\( lcmvalue \\geq 2 \\) and \\( emptiness \\geq 1 \\). It follows easily from [He] that up to the action of \\( \\mathrm{GL}_{lcmvalue}(\\mathbb{Z}) \\) and translation by lattice points, there are only finitely many convex lattice polytopes in \\( \\mathbb{R}^{lcmvalue} \\) having exactly \\( emptiness \\) interior lattice points. For \\( lcmvalue =2 \\) and \\( emptiness =1 \\), we get 16 polygons; see [PRV] for a figure showing all of them. Five of these 16 are triangles, and checking each case gives another proof that \\( |centerpoint\\ boundarypoint| /|boundarypoint\\ corepoint| \\leq 5 \\), and that equality is possible for only one of the five triangles."
    },
    "garbled_string": {
      "map": {
        "A": "qzxwvtnp",
        "B": "hjgrksla",
        "C": "vbcmdfni",
        "D": "lkjhtrew",
        "E": "asdfghjk",
        "F": "poiuytre",
        "M": "nmcbvzas",
        "P": "qwertyui",
        "x": "plmoknji",
        "x_1": "qazwsxed",
        "x_2": "edcrfvtg",
        "x_3": "tgbnhyuj",
        "y": "okmijnuh",
        "y_1": "yhnujmko",
        "y_2": "ujmikolp",
        "y_3": "ikolpjuj",
        "i": "polkmijn",
        "b": "mnbvcxza",
        "d": "qxcvbnma",
        "e": "fghjklqw",
        "n": "qwerasdf",
        "k": "zxcvrewq",
        "R": "poiulkjh"
      },
      "question": "which are collinear) have integer coordinates must have area greater than\nor equal to 5/2.",
      "solution": "Solution. A lattice polygon is a plane polygon whose vertices are lattice points, i.e., points with integer coordinates. The area of any convex lattice polygon has area equal to half an integer: this follows from Pick's Theorem mentioned in the second remark below; alternatively, by subdivision one can reduce to the case of a triangle, in which case the statement follows from the first remark below.\n\nConsider a convex lattice pentagon \\( qzxwvtnp hjgrksla vbcmdfni lkjhtrew asdfghjk \\) of minimal area. Since the area is always half an integer, the minimum exists. If the interior of side \\( qzxwvtnp hjgrksla \\) contains a lattice point \\( poiuytre \\), then \\( qzxwvtnp poiuytre vbcmdfni lkjhtrew asdfghjk \\) is a convex lattice pentagon with smaller area, contradicting the choice of \\( qzxwvtnp hjgrksla vbcmdfni lkjhtrew asdfghjk \\). (As is standard, vertices are listed in order around each polygon.) Applying this argument to each side, we may assume that all boundary lattice points are vertices.\n\nSeparate the vertices into four classes according to the parity of their coordinates. By the Pigeonhole Principle, one class must contain at least two vertices. The midpoint \\( nmcbvzas \\) between two such vertices has integer coordinates. By the previous paragraph, these two vertices cannot form a side of the polygon. Also, the pentagon is convex, so \\( nmcbvzas \\) is in the interior of the pentagon. Connecting \\( nmcbvzas \\) to the vertices divides the polygon into 5 triangles, each of area at least \\( 1 / 2 \\), so the whole polygon has area at least \\( 5 / 2 \\).\n\nRemark. The bound \\( 5 / 2 \\) cannot be improved: the polygon with vertices \\( (0,0) \\), \\( (1,0),(2,1),(1,2),(0,1) \\) is convex and has area \\( 5 / 2 \\) (see Figure 14).\n\nRemark. The area of a plane triangle with vertices \\( \\left(qazwsxed, yhnujmko\\right),\\left(edcrfvtg, ujmikolp\\right),\\left(tgbnhyuj, ikolpjuj\\right) \\) equals\n\\[\n\\frac{1}{2}\\left|\\operatorname{det}\\left(\\begin{array}{lll}\nqazwsxed & yhnujmko & 1 \\\\\nedcrfvtg & ujmikolp & 1 \\\\\ntgbnhyuj & ikolpjuj & 1\n\\end{array}\\right)\\right|\n\\]\n\nIn particular, if \\( qazwsxed, yhnujmko \\in \\mathbb{Z} \\), the area is half an integer.\nFor a formula for the volume of an \\( qwerasdf \\)-dimensional simplex in \\( \\mathbb{R}^{qwerasdf} \\), see Solution 5 to 1993B5.\n\nRemark (Pick's Theorem). Given a lattice polygon, let \\( polkmijn \\) be the number of internal lattice points and let \\( mnbvcxza \\) be the number of boundary lattice points. Pick's Theorem [Lar1, p. 68] states that the area of the polygon equals \\( polkmijn+mnbvcxza / 2-1 \\).\n\nIn the previous solution, we could have used Pick's Theorem in two places: in the first paragraph to prove that a lattice polygon has half-integer area (even if it is not convex), and as a substitute for the last sentence, using \\( polkmijn \\geq 1 \\) and \\( mnbvcxza=5 \\).\n\nRelated question. Problem 1981A6 [PutnamII, p. 37] is similar:\nSuppose that each of the vertices of \\( \\triangle qzxwvtnp hjgrksla vbcmdfni \\) is a lattice point in the \\( (plmoknji, okmijnuh) \\)-plane and that there is exactly one lattice point \\( qwertyui \\) in the interior of the triangle.\nThe line \\( qzxwvtnp qwertyui \\) is extended to meet \\( hjgrksla vbcmdfni \\) at \\( asdfghjk \\). Determine the largest possible value for the ratio of the lengths of segments \\( |qzxwvtnp qwertyui| /|qwertyui asdfghjk| \\).\nHere is a solution to 1981A6 using Pick's Theorem. (See [PutnamII, p. 118] for a non-Pick solution.)\n\nWe may reduce to the case where \\( hjgrksla vbcmdfni \\) has no lattice points on it other than \\( hjgrksla, vbcmdfni \\), and possibly \\( asdfghjk \\), by replacing the base \\( hjgrksla vbcmdfni \\) with the shortest segment along it with lattice endpoints and containing \\( asdfghjk \\) in its interior.\n\nCase 1: \\( asdfghjk \\) is a lattice point. Then the reflection of \\( asdfghjk \\) across \\( qwertyui \\) is also a lattice point, so it must coincide with \\( qzxwvtnp \\), and \\( |qzxwvtnp qwertyui| /|qwertyui asdfghjk|=1 \\).\n\nCase 2: \\( asdfghjk \\) is not a lattice point. Without loss of generality (by applying an affine transformation preserving the lattice), we may assume \\( hjgrksla=(0,0) \\) and \\( vbcmdfni=(1,0) \\). If \\( qzxwvtnp=(plmoknji, okmijnuh), okmijnuh>0 \\), then \\( plmoknji \\neq 0,1 \\), and by Pick's Theorem,\n\\[\n\\begin{aligned}\nokmijnuh & =2+\\#\\{\\text { boundary lattice points }\\}-2 \\\\\n& =qxcvbnma+fghjklqw+1\n\\end{aligned}\n\\]\nwhere \\( qxcvbnma=\\operatorname{gcd}(plmoknji, okmijnuh), fghjklqw=\\operatorname{gcd}(1-plmoknji, okmijnuh) \\). Since \\( qxcvbnma \\) and \\( fghjklqw \\) are relatively prime, qxcvbnma fghjklqw divides \\( okmijnuh \\), so \\( qxcvbnma fghjklqw \\leq qxcvbnma+fghjklqw+1 \\), or equivalently \\( (qxcvbnma-1)(fghjklqw-1) \\leq 2 \\). We have several subcases:\n- If \\( qxcvbnma=1 \\), then \\( okmijnuh=2+fghjklqw \\) is divisible by \\( fghjklqw \\), so \\( fghjklqw=1 \\) or 2 . If \\( fghjklqw=1 \\), then \\( okmijnuh=2 \\), and the ratio \\( |qzxwvtnp qwertyui| /|qwertyui asdfghjk| \\) is 2 . If \\( fghjklqw=2 \\), then \\( okmijnuh=3 \\), and the ratio is 3 .\n- If \\( fghjklqw=1 \\), then essentially the same argument gives a ratio of 2 or 3 .\n- The case \\( qxcvbnma=fghjklqw=2 \\) is not allowed, since \\( qxcvbnma \\) and \\( fghjklqw \\) are coprime.\n- If \\( qxcvbnma=3 \\) and \\( fghjklqw=2 \\), then \\( okmijnuh=6 \\), giving a ratio of 5 .\n- If \\( qxcvbnma=2 \\) and \\( fghjklqw=3 \\), then the ratio is also 5 .\n\nHence the maximum is 5 . This argument can be refined to show that equality is achieved only for one triangle (up to automorphisms of the plane preserving lattice points).\n\nRemark. The group of linear transformations of the plane preserving the lattice points and fixing the origin is the group \\( \\mathrm{GL}_{2}(\\mathbb{Z}) \\) defined in the introduction. It is important in number theory and related fields.\n\nRemark. Fix integers \\( qxcvbnma \\geq 2 \\) and \\( zxcvrewq \\geq 1 \\). It follows easily from [He] that up to the action of \\( \\mathrm{GL}_{qxcvbnma}(\\mathbb{Z}) \\) and translation by lattice points, there are only finitely many convex lattice polytopes in \\( \\mathbb{R}^{qwerasdf} \\) having exactly \\( zxcvrewq \\) interior lattice points. For \\( qxcvbnma=2 \\) and \\( zxcvrewq=1 \\), we get 16 polygons; see [PRV] for a figure showing all of them. Five of these 16 are triangles, and checking each case gives another proof that \\( |qzxwvtnp qwertyui| /|qwertyui asdfghjk| \\leq 5 \\), and that equality is possible for only one of the five triangles."
    },
    "kernel_variant": {
      "question": "Let a convex lattice hexagon be a convex polygon whose six vertices are lattice points (points with integer coordinates), with no three of the six vertices collinear.  Show that every such convex lattice hexagon has area at least $3$.\n\nMoreover, exhibit a convex lattice hexagon whose area is exactly $3$, thereby proving the bound is best possible.",
      "solution": "Write the vertices of the convex lattice hexagon P in counter‐clockwise order as A₁,A₂,A₃,A₄,A₅,A₆.\n\n1. (Half–integer areas.)  By the standard determinant formula or Pick’s Theorem, the area of any lattice polygon is a half‐integer.\n\n2. (No extra boundary points on a minimal hexagon.)  Suppose, for contradiction, that the theorem is false, and let P be a convex lattice hexagon of smallest possible (positive) area.  If some side AᵢAᵢ₊₁ contained an extra lattice point F strictly between Aᵢ and Aᵢ₊₁, then we form a new convex lattice hexagon P′ by replacing the two consecutive vertices …,Aᵢ,Aᵢ₊₁,… with …,Aᵢ,F,Aᵢ₊₂,… (that is, drop Aᵢ₊₁ and insert F in its place).  Because F lies on the boundary of P, the chord F Aᵢ₊₂ lies entirely inside P, so P′ is convex, has all vertices integral, and has strictly smaller area (we have removed the triangle AᵢAᵢ₊₁Aᵢ₊₂).  This contradicts the minimality of P.  Hence in a minimal‐area convex lattice hexagon each side contains no lattice point other than its two endpoints.\n\n3. (An interior lattice point.)  Lattice points fall into four parity classes ((even,even),(even,odd),(odd,even),(odd,odd)).  Among the six vertices A₁,…,A₆, two must lie in the same class by pigeonhole.  They cannot be adjacent, for then their midpoint would be a lattice point on that side, contradicting Step 2.  Thus we have two nonadjacent vertices, say Aᵢ and Aⱼ, that share parity.  Their midpoint M is therefore a lattice point, and since AᵢAⱼ is not a side, convexity forces M into the interior of P.\n\n4. (Triangulate about M.)  Join M to each vertex Aₖ.  This divides P into six triangles M A₁A₂, M A₂A₃,…,M A₆A₁.  None is degenerate (since M is not on any boundary edge), and each must have area ≥½ (every nonzero lattice‐triangle area is a positive half‐integer).\n\n5. (Lower bound on area.)  Summing,\n   area(P) = ∑ₖ area(M AₖAₖ₊₁) ≥ 6·½ = 3.\nSince the total area is also a half‐integer, the least it can be is 3.\n\n6. (Sharpness.)  The hexagon with vertices\n   (0,0), (1,0), (2,1), (2,2), (1,2), (0,1)\nin counter‐clockwise order is easily checked to be convex with no extra boundary points.  A direct shoelace or Pick computation gives\n   area = ½[(0·0+1·1+2·2+2·2+1·1+0·0) − (0·1+0·2+1·2+2·1+2·0+1·0)]\n         = ½(10 − 4) = 3.\n\nTherefore every convex lattice hexagon has area ≥3, and the example above shows 3 is best possible.",
      "_meta": {
        "core_steps": [
          "Lattice polygon areas are half-integers (via Pick or determinant).",
          "A minimum-area polygon can have no extra lattice points on its edges.",
          "With 5 vertices, pigeonhole on the 4 parity classes ⇒ two non-adjacent vertices share parity; their midpoint is an interior lattice point.",
          "Joining this midpoint to all vertices partitions the polygon into 5 triangles.",
          "Each triangle has area ≥ 1⁄2 ⇒ total area ≥ 5⁄2."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Number of vertices of the convex lattice polygon (currently a pentagon). Any value ≥5 keeps the pigeonhole step valid.",
            "original": 5
          },
          "slot2": {
            "description": "Resulting lower-bound on the area, equal to (number of vertices)/2.",
            "original": "5/2"
          },
          "slot3": {
            "description": "Concrete set of vertices used to show sharpness; any translation/rotation or other lattice pentagon of area 5/2 would work.",
            "original": "[(0,0), (1,0), (2,1), (1,2), (0,1)]"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}