{
  "index": "1953-A-2",
  "type": "COMB",
  "tag": [
    "COMB"
  ],
  "difficulty": "",
  "question": "2. Six points are in general position in space (no three in a line, no four in a plane). The fifteen line segments joining them in pairs are drawn and then painted, some segments red, some blue. Prove that some triangle has all its sides the same color.",
  "solution": "Solution. Let \\( P \\) be any one of the six points. Five of the line segments end at \\( P \\), and of these at least three, say \\( P Q, P R \\), and \\( P S \\), must have the same color, say blue. Then, if any one of the segments \\( Q R, R S, S Q \\) is blue we will have a blue triangle, and if not \\( Q R S \\) will be a red triangle. Thus in any event at least one triangle has all its sides the same color.\n\nRemarks. This problem later appeared as Problem E 1321, American Mathematical Monthly, vol. 66 (1959), pages 141-142. A. W. Goodman, \"On Sets of Acquaintances and Strangers at Any Party,\" American Mathematical Monthly, vol. 66 (1959), pages 778-783, shows that there must always be at least two monochromatic triangles. G. J. Simmons, in \"The Game of Sim,\" Journal of Recreational Mathematics, vol. 2 (1969), page 66, proposed the following game for two players.\n\nPlayers alternately color the segments red and blue, the object being to avoid making a triangle in one's own color. Since a monochromatic triangle must eventually be formed, this game cannot end in a draw. It has been shown that the second player can force a win. See Rounds and Yau, \"A Winning Strategy for Sim,\" Journal of Recreational Math., vol. 7\n(1974), pages 193-202; or Mead, Rosa, and Huang, \"The Game of Sim: A Winning Strategy for the Second Player,\" Mathematics Magazine, vol. 47 (1974), pages 243-247.\n\nGeneralizations to more points (and hence more line segments) and/or more colors have attracted considerable attention. For example, if the segments connecting 18 points are colored with two colors, there must be a monochromatic tetrahedron, but for 17 points there need not be such a tetrahedron. See A. M. Gleason and R. E. Greenwood, \"Combinatorial Relations and Chromatic Graphs,\" Canadian Journal of Mathematics, vol. 7 (1955), pages 1-7. A very general existence theorem along these lines was proved by F. P. Ramsey, \"On a Problem in Formal Logic,\" Proceedings of the London Mathematical Society, Ser. 2, vol. 30 (1930), pages 264-286; hence the least numbers which guarantee the existence of certain configurations are frequently called \"Ramsey numbers.\"\n\nSee also J. G. Kalbfleisch, \"Upper Bounds for Some Ramsey Numbers,\" Journal of Combinatorial Theory, vol. 2 (1967), pages 35-42; Keith Walker, \"Dichromatic Graphs and Ramsey Numbers,\" Journal of Combinatorial Theory, vol. 5 (1968), pages 238-243; and J. Spencer, \"Ramsey's Theorem -A New Lower Bound,\" Journal of Combinatorial Theory, vol. 18 (1975), pages 108-115.",
  "vars": [
    "P",
    "Q",
    "R",
    "S"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "P": "pointa",
        "Q": "pointb",
        "R": "pointc",
        "S": "pointd"
      },
      "question": "2. Six points are in general position in space (no three in a line, no four in a plane). The fifteen line segments joining them in pairs are drawn and then painted, some segments red, some blue. Prove that some triangle has all its sides the same color.",
      "solution": "Solution. Let \\( pointa \\) be any one of the six points. Five of the line segments end at \\( pointa \\), and of these at least three, say \\( pointa pointb, pointa pointc \\), and \\( pointa pointd \\), must have the same color, say blue. Then, if any one of the segments \\( pointb pointc, pointc pointd, pointd pointb \\) is blue we will have a blue triangle, and if not \\( pointb pointc pointd \\) will be a red triangle. Thus in any event at least one triangle has all its sides the same color.\n\nRemarks. This problem later appeared as Problem E 1321, American Mathematical Monthly, vol. 66 (1959), pages 141-142. A. W. Goodman, \"On Sets of Acquaintances and Strangers at Any Party,\" American Mathematical Monthly, vol. 66 (1959), pages 778-783, shows that there must always be at least two monochromatic triangles. G. J. Simmons, in \"The Game of Sim,\" Journal of Recreational Mathematics, vol. 2 (1969), page 66, proposed the following game for two players.\n\nPlayers alternately color the segments red and blue, the object being to avoid making a triangle in one's own color. Since a monochromatic triangle must eventually be formed, this game cannot end in a draw. It has been shown that the second player can force a win. See Rounds and Yau, \"A Winning Strategy for Sim,\" Journal of Recreational Math., vol. 7\n(1974), pages 193-202; or Mead, Rosa, and Huang, \"The Game of Sim: A Winning Strategy for the Second Player,\" Mathematics Magazine, vol. 47 (1974), pages 243-247.\n\nGeneralizations to more points (and hence more line segments) and/or more colors have attracted considerable attention. For example, if the segments connecting 18 points are colored with two colors, there must be a monochromatic tetrahedron, but for 17 points there need not be such a tetrahedron. See A. M. Gleason and pointc. E. Greenwood, \"Combinatorial Relations and Chromatic Graphs,\" Canadian Journal of Mathematics, vol. 7 (1955), pages 1-7. A very general existence theorem along these lines was proved by F. pointa. Ramsey, \"On a Problem in Formal Logic,\" Proceedings of the London Mathematical Society, Ser. 2, vol. 30 (1930), pages 264-286; hence the least numbers which guarantee the existence of certain configurations are frequently called \"Ramsey numbers.\"\n\nSee also J. G. Kalbfleisch, \"Upper Bounds for Some Ramsey Numbers,\" Journal of Combinatorial Theory, vol. 2 (1967), pages 35-42; Keith Walker, \"Dichromatic Graphs and Ramsey Numbers,\" Journal of Combinatorial Theory, vol. 5 (1968), pages 238-243; and J. Spencer, \"Ramsey's Theorem -A New Lower Bound,\" Journal of Combinatorial Theory, vol. 18 (1975), pages 108-115."
    },
    "descriptive_long_confusing": {
      "map": {
        "P": "sunflower",
        "Q": "bicycle",
        "R": "lemonade",
        "S": "umbrella"
      },
      "question": "2. Six points are in general position in space (no three in a line, no four in a plane). The fifteen line segments joining them in pairs are drawn and then painted, some segments red, some blue. Prove that some triangle has all its sides the same color.",
      "solution": "Solution. Let \\( sunflower \\) be any one of the six points. Five of the line segments end at \\( sunflower \\), and of these at least three, say \\( sunflower bicycle, sunflower lemonade \\), and \\( sunflower umbrella \\), must have the same color, say blue. Then, if any one of the segments \\( bicycle lemonade, lemonade umbrella, umbrella bicycle \\) is blue we will have a blue triangle, and if not \\( bicycle lemonade umbrella \\) will be a red triangle. Thus in any event at least one triangle has all its sides the same color.\n\nRemarks. This problem later appeared as Problem E 1321, American Mathematical Monthly, vol. 66 (1959), pages 141-142. A. W. Goodman, \"On Sets of Acquaintances and Strangers at Any Party,\" American Mathematical Monthly, vol. 66 (1959), pages 778-783, shows that there must always be at least two monochromatic triangles. G. J. Simmons, in \"The Game of Sim,\" Journal of Recreational Mathematics, vol. 2 (1969), page 66, proposed the following game for two players.\n\nPlayers alternately color the segments red and blue, the object being to avoid making a triangle in one's own color. Since a monochromatic triangle must eventually be formed, this game cannot end in a draw. It has been shown that the second player can force a win. See Rounds and Yau, \"A Winning Strategy for Sim,\" Journal of Recreational Math., vol. 7\n(1974), pages 193-202; or Mead, Rosa, and Huang, \"The Game of Sim: A Winning Strategy for the Second Player,\" Mathematics Magazine, vol. 47 (1974), pages 243-247.\n\nGeneralizations to more points (and hence more line segments) and/or more colors have attracted considerable attention. For example, if the segments connecting 18 points are colored with two colors, there must be a monochromatic tetrahedron, but for 17 points there need not be such a tetrahedron. See A. M. Gleason and R. E. Greenwood, \"Combinatorial Relations and Chromatic Graphs,\" Canadian Journal of Mathematics, vol. 7 (1955), pages 1-7. A very general existence theorem along these lines was proved by F. P. Ramsey, \"On a Problem in Formal Logic,\" Proceedings of the London Mathematical Society, Ser. 2, vol. 30 (1930), pages 264-286; hence the least numbers which guarantee the existence of certain configurations are frequently called \"Ramsey numbers.\"\n\nSee also J. G. Kalbfleisch, \"Upper Bounds for Some Ramsey Numbers,\" Journal of Combinatorial Theory, vol. 2 (1967), pages 35-42; Keith Walker, \"Dichromatic Graphs and Ramsey Numbers,\" Journal of Combinatorial Theory, vol. 5 (1968), pages 238-243; and J. Spencer, \"Ramsey's Theorem -A New Lower Bound,\" Journal of Combinatorial Theory, vol. 18 (1975), pages 108-115."
    },
    "descriptive_long_misleading": {
      "map": {
        "P": "voidspace",
        "Q": "abysspoint",
        "R": "nothingness",
        "S": "vacuumsite"
      },
      "question": "Problem:\n<<<\n2. Six points are in general position in space (no three in a line, no four in a plane). The fifteen line segments joining them in pairs are drawn and then painted, some segments red, some blue. Prove that some triangle has all its sides the same color.\n>>>\n",
      "solution": "Solution:\n<<<\nSolution. Let \\( voidspace \\) be any one of the six points. Five of the line segments end at \\( voidspace \\), and of these at least three, say \\( voidspace abysspoint, voidspace nothingness \\), and \\( voidspace vacuumsite \\), must have the same color, say blue. Then, if any one of the segments \\( abysspoint nothingness, nothingness vacuumsite, vacuumsite abysspoint \\) is blue we will have a blue triangle, and if not \\( abysspoint nothingness vacuumsite \\) will be a red triangle. Thus in any event at least one triangle has all its sides the same color.\n\nRemarks. This problem later appeared as Problem E 1321, American Mathematical Monthly, vol. 66 (1959), pages 141-142. A. W. Goodman, \"On Sets of Acquaintances and Strangers at Any Party,\" American Mathematical Monthly, vol. 66 (1959), pages 778-783, shows that there must always be at least two monochromatic triangles. G. J. Simmons, in \"The Game of Sim,\" Journal of Recreational Mathematics, vol. 2 (1969), page 66, proposed the following game for two players.\n\nPlayers alternately color the segments red and blue, the object being to avoid making a triangle in one's own color. Since a monochromatic triangle must eventually be formed, this game cannot end in a draw. It has been shown that the second player can force a win. See Rounds and Yau, \"A Winning Strategy for Sim,\" Journal of Recreational Math., vol. 7\n(1974), pages 193-202; or Mead, Rosa, and Huang, \"The Game of Sim: A Winning Strategy for the Second Player,\" Mathematics Magazine, vol. 47 (1974), pages 243-247.\n\nGeneralizations to more points (and hence more line segments) and/or more colors have attracted considerable attention. For example, if the segments connecting 18 points are colored with two colors, there must be a monochromatic tetrahedron, but for 17 points there need not be such a tetrahedron. See A. M. Gleason and R. E. Greenwood, \"Combinatorial Relations and Chromatic Graphs,\" Canadian Journal of Mathematics, vol. 7 (1955), pages 1-7. A very general existence theorem along these lines was proved by F. P. Ramsey, \"On a Problem in Formal Logic,\" Proceedings of the London Mathematical Society, Ser. 2, vol. 30 (1930), pages 264-286; hence the least numbers which guarantee the existence of certain configurations are frequently called \"Ramsey numbers.\"\n\nSee also J. G. Kalbfleisch, \"Upper Bounds for Some Ramsey Numbers,\" Journal of Combinatorial Theory, vol. 2 (1967), pages 35-42; Keith Walker, \"Dichromatic Graphs and Ramsey Numbers,\" Journal of Combinatorial Theory, vol. 5 (1968), pages 238-243; and J. Spencer, \"Ramsey's Theorem -A New Lower Bound,\" Journal of Combinatorial Theory, vol. 18 (1975), pages 108-115.\n>>>\n"
    },
    "garbled_string": {
      "map": {
        "P": "qzxwvtnp",
        "Q": "hjgrksla",
        "R": "mnbvcxqe",
        "S": "lkjhgfdp"
      },
      "question": "2. Six points are in general position in space (no three in a line, no four in a plane). The fifteen line segments joining them in pairs are drawn and then painted, some segments red, some blue. Prove that some triangle has all its sides the same color.",
      "solution": "Solution. Let \\( qzxwvtnp \\) be any one of the six points. Five of the line segments end at \\( qzxwvtnp \\), and of these at least three, say \\( qzxwvtnp hjgrksla, qzxwvtnp mnbvcxqe \\), and \\( qzxwvtnp lkjhgfdp \\), must have the same color, say blue. Then, if any one of the segments \\( hjgrksla mnbvcxqe, mnbvcxqe lkjhgfdp, lkjhgfdp hjgrksla \\) is blue we will have a blue triangle, and if not \\( hjgrksla mnbvcxqe lkjhgfdp \\) will be a red triangle. Thus in any event at least one triangle has all its sides the same color.\n\nRemarks. This problem later appeared as Problem E 1321, American Mathematical Monthly, vol. 66 (1959), pages 141-142. A. W. Goodman, \"On Sets of Acquaintances and Strangers at Any Party,\" American Mathematical Monthly, vol. 66 (1959), pages 778-783, shows that there must always be at least two monochromatic triangles. G. J. Simmons, in \"The Game of Sim,\" Journal of Recreational Mathematics, vol. 2 (1969), page 66, proposed the following game for two players.\n\nPlayers alternately color the segments red and blue, the object being to avoid making a triangle in one's own color. Since a monochromatic triangle must eventually be formed, this game cannot end in a draw. It has been shown that the second player can force a win. See Rounds and Yau, \"A Winning Strategy for Sim,\" Journal of Recreational Math., vol. 7\n(1974), pages 193-202; or Mead, Rosa, and Huang, \"The Game of Sim: A Winning Strategy for the Second Player,\" Mathematics Magazine, vol. 47 (1974), pages 243-247.\n\nGeneralizations to more points (and hence more line segments) and/or more colors have attracted considerable attention. For example, if the segments connecting 18 points are colored with two colors, there must be a monochromatic tetrahedron, but for 17 points there need not be such a tetrahedron. See A. M. Gleason and R. E. Greenwood, \"Combinatorial Relations and Chromatic Graphs,\" Canadian Journal of Mathematics, vol. 7 (1955), pages 1-7. A very general existence theorem along these lines was proved by F. P. Ramsey, \"On a Problem in Formal Logic,\" Proceedings of the London Mathematical Society, Ser. 2, vol. 30 (1930), pages 264-286; hence the least numbers which guarantee the existence of certain configurations are frequently called \"Ramsey numbers.\"\n\nSee also J. G. Kalbfleisch, \"Upper Bounds for Some Ramsey Numbers,\" Journal of Combinatorial Theory, vol. 2 (1967), pages 35-42; Keith Walker, \"Dichromatic Graphs and Ramsey Numbers,\" Journal of Combinatorial Theory, vol. 5 (1968), pages 238-243; and J. Spencer, \"Ramsey's Theorem -A New Lower Bound,\" Journal of Combinatorial Theory, vol. 18 (1975), pages 108-115."
    },
    "kernel_variant": {
      "question": "A countably infinite set S of points is placed in the Euclidean plane in general position (no three points are collinear).  \nFor every unordered pair of distinct points of S the segment joining them is drawn and then coloured, independently, Crimson, Jade, or Sapphire.  Prove that at least one of the following three monochromatic configurations must appear:\n\n(A) An infinite Crimson clique – an infinite subset of S in which every segment determined by two of its points is Crimson.\n\n(B) A Sapphire convex hexagon – six points that are the vertices of a strictly convex hexagon such that all fifteen of the segments joining these six points are Sapphire.\n\n(C) A Jade convex pentagon – five points that are the vertices of a strictly convex pentagon such that all ten of the segments determined by these five vertices are Jade.  (No condition is imposed on the interior of the pentagon.)",
      "solution": "Two classical theorems will be used repeatedly.\n\n(i) Infinite Ramsey theorem for three colours (Ramsey–Erdős–Rado).  \n Every 3-colouring of the edges of the complete graph on a countably infinite vertex set contains an infinite monochromatic complete subgraph.\n\n(ii) Erdős–Szekeres “Happy–Ending” theorem.  \n For every integer r ≥ 3 there is a least integer ES(r) such that any ES(r) points in the plane in general position contain r points in strictly convex position.  It is known that ES(5)=10 and ES(6)=17.\n\nStep 1.  Apply (i) to the 3-colouring of the segments of S.  We obtain an infinite subset  \n T ⊆ S whose every connecting segment has the same colour κ ∈ {Crimson, Jade, Sapphire}.\n\nStep 2.  Distinguish three cases according to κ.\n\n• Case κ = Crimson.  \n  Then T itself is an infinite Crimson clique, so configuration (A) occurs.\n\n• Case κ = Sapphire.  \n  By (ii) with r = 6, every set of 17 points in general position contains six in strictly convex position.  Select 17 arbitrary points of T; the theorem supplies six of them, say P₁,…,P₆, in convex position.  Because all segments inside T are Sapphire, the fifteen segments PᵢPⱼ are Sapphire, giving configuration (B).\n\n• Case κ = Jade.  \n  Here every segment joining two points of T is Jade.  Use (ii) with r = 5: any 10 points in general position contain five in strictly convex position.  Taking 10 arbitrary points of T yields five such points, Q₁,…,Q₅.  All ten segments QᵢQⱼ (1 ≤ i < j ≤ 5) are therefore Jade, providing configuration (C).\n\nStep 3.  Exactly one of the foregoing possibilities for the monochromatic clique T occurs; in that eventuality the corresponding configuration (A), (B) or (C) has been produced.  Hence every 3-colouring of the segments determined by a countably infinite planar point-set in general position must contain at least one of the three stated monochromatic configurations. ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.453840",
        "was_fixed": false,
        "difficulty_analysis": "1.  Multiple interacting theories.  \n   • Infinite Ramsey theory (Erdős–Rado) is required to pass from a 3-coloring on a countable set to an infinite monochromatic clique.  \n   • The Erdős–Szekeres theorem is then invoked to extract large convex polygons from infinite monochromatic point sets.  \n   • Harborth’s Empty-Pentagon Theorem supplies a non-trivial geometric refinement (the 5-hole) that does not follow from Erdős–Szekeres alone.\n\n2.  Higher conceptual load.  \n   The solver must be comfortable moving among discrete infinite combinatorics, classical convex-geometry extremal results, and “empty-polygon” theory—far beyond the pigeon-hole argument that resolves the original six-point problem.\n\n3.  Cascading case analysis.  \n   Successive reductions (first to an infinite monochromatic clique, then to specific convex-geometric structures) make the logical structure deeper and require careful coordination of results drawn from different areas.\n\n4.  Non-constructive existence theorems.  \n   All three outside theorems used (infinite Ramsey, Erdős–Szekeres, Harborth) are existential and non-constructive, demanding a higher level of theoretical sophistication than the original one-page proof.\n\nBecause the enhanced variant forces the competitor to blend infinite combinatorics with advanced planar geometry and to handle several substantial theorems, it is substantially harder than both the original problem and the previous kernel variant while still remaining well-posed and solvable."
      }
    },
    "original_kernel_variant": {
      "question": "A countably infinite set S of points is placed in the Euclidean plane in general position (no three points are collinear).  \nFor every unordered pair of distinct points of S the segment joining them is drawn and then coloured, independently, Crimson, Jade, or Sapphire.  Prove that at least one of the following three monochromatic configurations must appear:\n\n(A) An infinite Crimson clique – an infinite subset of S in which every segment determined by two of its points is Crimson.\n\n(B) A Sapphire convex hexagon – six points that are the vertices of a strictly convex hexagon such that all fifteen of the segments joining these six points are Sapphire.\n\n(C) A Jade convex pentagon – five points that are the vertices of a strictly convex pentagon such that all ten of the segments determined by these five vertices are Jade.  (No condition is imposed on the interior of the pentagon.)",
      "solution": "Two classical theorems will be used repeatedly.\n\n(i) Infinite Ramsey theorem for three colours (Ramsey–Erdős–Rado).  \n Every 3-colouring of the edges of the complete graph on a countably infinite vertex set contains an infinite monochromatic complete subgraph.\n\n(ii) Erdős–Szekeres “Happy–Ending” theorem.  \n For every integer r ≥ 3 there is a least integer ES(r) such that any ES(r) points in the plane in general position contain r points in strictly convex position.  It is known that ES(5)=10 and ES(6)=17.\n\nStep 1.  Apply (i) to the 3-colouring of the segments of S.  We obtain an infinite subset  \n T ⊆ S whose every connecting segment has the same colour κ ∈ {Crimson, Jade, Sapphire}.\n\nStep 2.  Distinguish three cases according to κ.\n\n• Case κ = Crimson.  \n  Then T itself is an infinite Crimson clique, so configuration (A) occurs.\n\n• Case κ = Sapphire.  \n  By (ii) with r = 6, every set of 17 points in general position contains six in strictly convex position.  Select 17 arbitrary points of T; the theorem supplies six of them, say P₁,…,P₆, in convex position.  Because all segments inside T are Sapphire, the fifteen segments PᵢPⱼ are Sapphire, giving configuration (B).\n\n• Case κ = Jade.  \n  Here every segment joining two points of T is Jade.  Use (ii) with r = 5: any 10 points in general position contain five in strictly convex position.  Taking 10 arbitrary points of T yields five such points, Q₁,…,Q₅.  All ten segments QᵢQⱼ (1 ≤ i < j ≤ 5) are therefore Jade, providing configuration (C).\n\nStep 3.  Exactly one of the foregoing possibilities for the monochromatic clique T occurs; in that eventuality the corresponding configuration (A), (B) or (C) has been produced.  Hence every 3-colouring of the segments determined by a countably infinite planar point-set in general position must contain at least one of the three stated monochromatic configurations. ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.388315",
        "was_fixed": false,
        "difficulty_analysis": "1.  Multiple interacting theories.  \n   • Infinite Ramsey theory (Erdős–Rado) is required to pass from a 3-coloring on a countable set to an infinite monochromatic clique.  \n   • The Erdős–Szekeres theorem is then invoked to extract large convex polygons from infinite monochromatic point sets.  \n   • Harborth’s Empty-Pentagon Theorem supplies a non-trivial geometric refinement (the 5-hole) that does not follow from Erdős–Szekeres alone.\n\n2.  Higher conceptual load.  \n   The solver must be comfortable moving among discrete infinite combinatorics, classical convex-geometry extremal results, and “empty-polygon” theory—far beyond the pigeon-hole argument that resolves the original six-point problem.\n\n3.  Cascading case analysis.  \n   Successive reductions (first to an infinite monochromatic clique, then to specific convex-geometric structures) make the logical structure deeper and require careful coordination of results drawn from different areas.\n\n4.  Non-constructive existence theorems.  \n   All three outside theorems used (infinite Ramsey, Erdős–Szekeres, Harborth) are existential and non-constructive, demanding a higher level of theoretical sophistication than the original one-page proof.\n\nBecause the enhanced variant forces the competitor to blend infinite combinatorics with advanced planar geometry and to handle several substantial theorems, it is substantially harder than both the original problem and the previous kernel variant while still remaining well-posed and solvable."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}