{
  "index": "1962-A-1",
  "type": "GEO",
  "tag": [
    "GEO",
    "COMB"
  ],
  "difficulty": "",
  "question": "1. Given five points in a plane, no three of which lie on a straight line, show that some four of these points form the vertices of a convex quadrilateral.",
  "solution": "Solution. Consider the convex hull \\( K \\) of the five points. If \\( K \\) is a pentagon or a quadrilateral, the result is obvious. Since the points do not lie on a line, the only remaining possibility is that \\( K \\) is a triangle \\( A B C \\) with vertices among the given points. The other two points \\( D \\) and \\( E \\) are in the interior of \\( A B C \\) and the line \\( D E \\) meets the triangle twice, not at a vertex. By relettering if necessary, we may suppose that \\( D E \\) meets \\( A B \\) and \\( A C \\) but not \\( B C \\). Then \\( B, C, D \\), and \\( E \\) are the vertices of a convex quadrilateral.\n\nRemark. This problem and its generalizations to more points were first studied by Esther Klein Szekeres. A proof that every set of nine points with no three collinear contains the vertices of a convex pentagon, and further references, will be found in W. E. Bonnice, \"On Convex Planar Polygons Determined by a Finite Planar Set,\" American Mathematical Monthly, vol. 81 (1974), pages 749-752.",
  "vars": [
    "K",
    "A",
    "B",
    "C",
    "D",
    "E"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "K": "hullregion",
        "A": "vertexalpha",
        "B": "vertexbeta",
        "C": "vertexgamma",
        "D": "vertexdelta",
        "E": "vertexepsilon"
      },
      "question": "1. Given five points in a plane, no three of which lie on a straight line, show that some four of these points form the vertices of a convex quadrilateral.",
      "solution": "Solution. Consider the convex hull \\( hullregion \\) of the five points. If \\( hullregion \\) is a pentagon or a quadrilateral, the result is obvious. Since the points do not lie on a line, the only remaining possibility is that \\( hullregion \\) is a triangle \\( vertexalpha vertexbeta vertexgamma \\) with vertices among the given points. The other two points \\( vertexdelta \\) and \\( vertexepsilon \\) are in the interior of \\( vertexalpha vertexbeta vertexgamma \\) and the line \\( vertexdelta vertexepsilon \\) meets the triangle twice, not at a vertex. By relettering if necessary, we may suppose that \\( vertexdelta vertexepsilon \\) meets \\( vertexalpha vertexbeta \\) and \\( vertexalpha vertexgamma \\) but not \\( vertexbeta vertexgamma \\). Then \\( vertexbeta, vertexgamma, vertexdelta \\), and \\( vertexepsilon \\) are the vertices of a convex quadrilateral.\n\nRemark. This problem and its generalizations to more points were first studied by Esther Klein Szekeres. A proof that every set of nine points with no three collinear contains the vertices of a convex pentagon, and further references, will be found in W. E. Bonnice, \"On Convex Planar Polygons Determined by a Finite Planar Set,\" American Mathematical Monthly, vol. 81 (1974), pages 749-752."
    },
    "descriptive_long_confusing": {
      "map": {
        "K": "blueprint",
        "A": "pineapple",
        "B": "harmonica",
        "C": "sandpaper",
        "D": "buttercup",
        "E": "clockwork"
      },
      "question": "1. Given five points in a plane, no three of which lie on a straight line, show that some four of these points form the vertices of a convex quadrilateral.",
      "solution": "Solution. Consider the convex hull \\( blueprint \\) of the five points. If \\( blueprint \\) is a pentagon or a quadrilateral, the result is obvious. Since the points do not lie on a line, the only remaining possibility is that \\( blueprint \\) is a triangle \\( pineapple harmonica sandpaper \\) with vertices among the given points. The other two points \\( buttercup \\) and \\( clockwork \\) are in the interior of \\( pineapple harmonica sandpaper \\) and the line \\( buttercup clockwork \\) meets the triangle twice, not at a vertex. By relettering if necessary, we may suppose that \\( buttercup clockwork \\) meets \\( pineapple harmonica \\) and \\( pineapple sandpaper \\) but not \\( harmonica sandpaper \\). Then \\( harmonica, sandpaper, buttercup \\), and \\( clockwork \\) are the vertices of a convex quadrilateral.\n\nRemark. This problem and its generalizations to more points were first studied by Esther Klein Szekeres. A proof that every set of nine points with no three collinear contains the vertices of a convex pentagon, and further references, will be found in W. E. Bonnice, \"On Convex Planar Polygons Determined by a Finite Planar Set,\" American Mathematical Monthly, vol. 81 (1974), pages 749-752."
    },
    "descriptive_long_misleading": {
      "map": {
        "K": "concaveregion",
        "A": "centerpoint",
        "B": "collinearpoint",
        "C": "edgepoint",
        "D": "boundarypoint",
        "E": "surfacepoint"
      },
      "question": "1. Given five points in a plane, no three of which lie on a straight line, show that some four of these points form the vertices of a convex quadrilateral.",
      "solution": "Solution. Consider the convex hull \\( concaveregion \\) of the five points. If \\( concaveregion \\) is a pentagon or a quadrilateral, the result is obvious. Since the points do not lie on a line, the only remaining possibility is that \\( concaveregion \\) is a triangle \\( centerpoint\\, collinearpoint\\, edgepoint \\) with vertices among the given points. The other two points \\( boundarypoint \\) and \\( surfacepoint \\) are in the interior of \\( centerpoint\\, collinearpoint\\, edgepoint \\) and the line \\( boundarypoint\\, surfacepoint \\) meets the triangle twice, not at a vertex. By relettering if necessary, we may suppose that \\( boundarypoint\\, surfacepoint \\) meets \\( centerpoint\\, collinearpoint \\) and \\( centerpoint\\, edgepoint \\) but not \\( collinearpoint\\, edgepoint \\). Then \\( collinearpoint, edgepoint, boundarypoint \\), and \\( surfacepoint \\) are the vertices of a convex quadrilateral.\n\nRemark. This problem and its generalizations to more points were first studied by Esther Klein Szekeres. A proof that every set of nine points with no three collinear contains the vertices of a convex pentagon, and further references, will be found in W. E. Bonnice, \"On Convex Planar Polygons Determined by a Finite Planar Set,\" American Mathematical Monthly, vol. 81 (1974), pages 749-752."
    },
    "garbled_string": {
      "map": {
        "K": "qzxwvtnp",
        "A": "hjgrksla",
        "B": "xcmvafqd",
        "C": "eynpsrlo",
        "D": "uzkbtfia",
        "E": "wjodmcre"
      },
      "question": "1. Given five points in a plane, no three of which lie on a straight line, show that some four of these points form the vertices of a convex quadrilateral.",
      "solution": "Solution. Consider the convex hull \\( qzxwvtnp \\) of the five points. If \\( qzxwvtnp \\) is a pentagon or a quadrilateral, the result is obvious. Since the points do not lie on a line, the only remaining possibility is that \\( qzxwvtnp \\) is a triangle \\( hjgrksla xcmvafqd eynpsrlo \\) with vertices among the given points. The other two points \\( uzkbtfia \\) and \\( wjodmcre \\) are in the interior of \\( hjgrksla xcmvafqd eynpsrlo \\) and the line \\( uzkbtfia wjodmcre \\) meets the triangle twice, not at a vertex. By relettering if necessary, we may suppose that \\( uzkbtfia wjodmcre \\) meets \\( hjgrksla xcmvafqd \\) and \\( hjgrksla eynpsrlo \\) but not \\( xcmvafqd eynpsrlo \\). Then \\( xcmvafqd, eynpsrlo, uzkbtfia \\), and \\( wjodmcre \\) are the vertices of a convex quadrilateral.\n\nRemark. This problem and its generalizations to more points were first studied by Esther Klein Szekeres. A proof that every set of nine points with no three collinear contains the vertices of a convex pentagon, and further references, will be found in W. E. Bonnice, \"On Convex Planar Polygons Determined by a Finite Planar Set,\" American Mathematical Monthly, vol. 81 (1974), pages 749-752."
    },
    "kernel_variant": {
      "question": "Let  S  be a set of six points in the Euclidean plane, no three of which are collinear.  Prove that four points of  S  are the vertices of a convex quadrilateral.",
      "solution": "We prove the assertion by distinguishing the possible shapes of the convex hull  H  of the six points.\n\n--------------------------------------------------\nLemma (Erdős–Szekeres, 1935 –  5-point version).\nEvery set of five points in the plane, no three collinear, contains four that are the vertices of a convex quadrilateral.\n\nSketch of proof of the lemma.\nLabel the five points so that their x–coordinates increase:\nP₁, P₂, … , P₅.\nLook at the y–coordinates y₁ ,…, y₅. \nBy the classical Erdős–Szekeres monotone-subsequence lemma, among five real numbers there is an increasing or a decreasing subsequence of length three.  Hence there are indices i<j<k with either yᵢ<yⱼ<y_k (call such a triple a ‘cup’) or yᵢ>yⱼ>y_k (a ‘cap’).\nIf {Pᵢ ,Pⱼ ,P_k } is a cup, then together with the left-most of the remaining two points they form the vertices of a convex quadrilateral; if it is a cap, the same holds with the right-most remaining point.  (Because the x–coordinates are strictly ordered, the four points appear in cyclic order around their convex hull and no one lies inside the triangle formed by the other three.)  □\n--------------------------------------------------\n\nNow return to the original six-point set  S.\n\nCase 1.  The convex hull  H  of  S  has at least four vertices.\nThe hull can then only be a quadrilateral, a pentagon or a hexagon.  In any of these situations any four consecutive hull-vertices already form the required convex quadrilateral.\n\nCase 2.  The convex hull  H  is a triangle, say  △ABC.  All other three points of  S  lie strictly inside this triangle.  Remove one hull vertex, say  A.  The remaining five points\n   T = S \\ {A}\nform a five-point set in general position, so by the lemma just proved there is a convex quadrilateral whose vertices are four points of  T, hence four points of the original set  S.\n\nSince one of the two cases must occur, every six-point set with no three collinear contains a convex quadrilateral.",
      "_meta": {
        "core_steps": [
          "Take the convex hull of the point set.",
          "If the hull already has ≥4 vertices, those vertices give a convex quadrilateral.",
          "Otherwise the hull is a triangle that contains at least two other given points in its interior.",
          "Join two interior points; the segment meets two different sides of the triangle (never at a vertex).",
          "The two intersection‐side vertices together with the two interior points form a convex quadrilateral."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Total number of given points (must only guarantee at least two points inside a possible triangular hull).",
            "original": "5"
          },
          "slot2": {
            "description": "Explicit list of possible convex‐hull shapes with ≥4 vertices (here named ‘pentagon or quadrilateral’; for more points it could be ‘hexagon, heptagon, …’).",
            "original": "“pentagon or a quadrilateral”"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}