{
  "index": "2000-B-6",
  "type": "COMB",
  "tag": [
    "COMB",
    "GEO"
  ],
  "difficulty": "",
  "question": "Let $B$ be a set of more than $2^{n+1}/n$ distinct points with\ncoordinates\nof the form $(\\pm 1,\\pm 1,\\ldots,\\pm 1)$ in $n$-dimensional space with\n$n\\geq 3$.\nShow that there are three distinct points in $B$ which are the vertices of\nan\nequilateral triangle.\n\n\\end{itemize}\n\\end{document}",
  "solution": "For each point $P$ in $B$, let $S_P$ be the set of points with\nall coordinates equal to $\\pm 1$ which\ndiffer from $P$ in exactly one coordinate. Since there are more than\n$2^{n+1}/n$ points in $B$, and each $S_P$ has $n$ elements, the\ncardinalities of the sets $S_P$ add up to more than $2^{n+1}$, which\nis to say, more than twice the total number of points. By the\npigeonhole principle, there must be a point in three of the\nsets, say $S_P, S_Q, S_R$. But then any two of $P, Q, R$ differ in\nexactly two coordinates, so $PQR$ is an equilateral triangle, as\ndesired.\n\n\n\\end{itemize}\n\n\\end{document}",
  "vars": [
    "P",
    "Q",
    "R",
    "S_P",
    "S_Q",
    "S_R"
  ],
  "params": [
    "B",
    "n"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "P": "vertexone",
        "Q": "vertextwo",
        "R": "vertexthr",
        "S_P": "adjacentp",
        "S_Q": "adjacentq",
        "S_R": "adjacentr",
        "B": "pointset",
        "n": "dimension"
      },
      "question": "Let $pointset$ be a set of more than $2^{dimension+1}/dimension$ distinct points with\ncoordinates\nof the form $(\\pm 1,\\pm 1,\\ldots,\\pm 1)$ in $dimension$-dimensional space with\n$dimension\\geq 3$.\nShow that there are three distinct points in $pointset$ which are the vertices of\nan\nequilateral triangle.\n\n\\end{itemize}\n\\end{document}",
      "solution": "For each point $vertexone$ in $pointset$, let $adjacentp$ be the set of points with\nall coordinates equal to $\\pm 1$ which\ndiffer from $vertexone$ in exactly one coordinate. Since there are more than\n$2^{dimension+1}/dimension$ points in $pointset$, and each $adjacentp$ has $dimension$ elements, the\ncardinalities of the sets $adjacentp$ add up to more than $2^{dimension+1}$, which\nis to say, more than twice the total number of points. By the\npigeonhole principle, there must be a point in three of the\nsets, say $adjacentp, adjacentq, adjacentr$. But then any two of $vertexone, vertextwo, vertexthr$ differ in\nexactly two coordinates, so $vertexonevertextwovertexthr$ is an equilateral triangle, as\ndesired.\n\n\n\\end{itemize}\n\n\\end{document}"
    },
    "descriptive_long_confusing": {
      "map": {
        "P": "pinecone",
        "Q": "quagmire",
        "R": "raincloud",
        "S_P": "treadmill",
        "S_Q": "sandstorm",
        "S_R": "palmolive",
        "B": "blackbird",
        "n": "narrowness"
      },
      "question": "\nLet $blackbird$ be a set of more than $2^{narrowness+1}/narrowness$ distinct points with\ncoordinates\nof the form $(\\pm 1,\\pm 1,\\ldots,\\pm 1)$ in $narrowness$-dimensional space with\n$narrowness\\geq 3$.\nShow that there are three distinct points in $blackbird$ which are the vertices of\nan\nequilateral triangle.\n\n\\end{itemize}\n\\end{document}\n",
      "solution": "\nFor each point $pinecone$ in $blackbird$, let $treadmill$ be the set of points with\nall coordinates equal to $\\pm 1$ which\ndiffer from $pinecone$ in exactly one coordinate. Since there are more than\n$2^{narrowness+1}/narrowness$ points in $blackbird$, and each $treadmill$ has $narrowness$ elements, the\ncardinalities of the sets $treadmill$ add up to more than $2^{narrowness+1}$, which\nis to say, more than twice the total number of points. By the\npigeonhole principle, there must be a point in three of the\nsets, say $treadmill, sandstorm, palmolive$. But then any two of $pinecone, quagmire, raincloud$ differ in\nexactly two coordinates, so $pinecone quagmire raincloud$ is an equilateral triangle, as\ndesired.\n\n\n\\end{itemize}\n\n\\end{document}\n"
    },
    "descriptive_long_misleading": {
      "map": {
        "P": "emptiness",
        "Q": "continuum",
        "R": "infinity",
        "S_P": "voidrealm",
        "S_Q": "boundless",
        "S_R": "silentsea",
        "B": "singleton",
        "n": "limitless"
      },
      "question": "Let $singleton$ be a set of more than $2^{limitless+1}/limitless$ distinct points with\ncoordinates\nof the form $(\\pm 1,\\pm 1,\\ldots,\\pm 1)$ in $limitless$-dimensional space with\n$limitless\\geq 3$.\nShow that there are three distinct points in $singleton$ which are the vertices of\nan\nequilateral triangle.",
      "solution": "For each point $emptiness$ in $singleton$, let $voidrealm$ be the set of points with\nall coordinates equal to $\\pm 1$ which\ndiffer from $emptiness$ in exactly one coordinate. Since there are more than\n$2^{limitless+1}/limitless$ points in $singleton$, and each $voidrealm$ has $limitless$ elements, the\ncardinalities of the sets $voidrealm$ add up to more than $2^{limitless+1}$, which\nis to say, more than twice the total number of points. By the\npigeonhole principle, there must be a point in three of the\nsets, say $voidrealm, boundless, silentsea$. But then any two of $emptiness, continuum, infinity$ differ in\nexactly two coordinates, so $emptinesscontinuuminfinity$ is an equilateral triangle, as\ndesired."
    },
    "garbled_string": {
      "map": {
        "P": "qzxwvtnp",
        "Q": "hjgrksla",
        "R": "mbvcqryu",
        "S_P": "lwkdmfja",
        "S_Q": "zkrptuav",
        "S_R": "nxyqrsop",
        "B": "fjdospwe",
        "n": "ghtlbrmq"
      },
      "question": "Let $fjdospwe$ be a set of more than $2^{ghtlbrmq+1}/ghtlbrmq$ distinct points with\ncoordinates\nof the form $(\\pm 1,\\pm 1,\\ldots,\\pm 1)$ in $ghtlbrmq$-dimensional space with\n$ghtlbrmq\\geq 3$.\nShow that there are three distinct points in $fjdospwe$ which are the vertices of\nan\nequilateral triangle.",
      "solution": "For each point $qzxwvtnp$ in $fjdospwe$, let $lwkdmfja$ be the set of points with\nall coordinates equal to $\\pm 1$ which\ndiffer from $qzxwvtnp$ in exactly one coordinate. Since there are more than\n$2^{ghtlbrmq+1}/ghtlbrmq$ points in $fjdospwe$, and each $lwkdmfja$ has $ghtlbrmq$ elements, the\ncardinalities of the sets $lwkdmfja$ add up to more than $2^{ghtlbrmq+1}$, which\nis to say, more than twice the total number of points. By the\npigeonhole principle, there must be a point in three of the\nsets, say $lwkdmfja, zkrptuav, nxyqrsop$. But then any two of $qzxwvtnp, hjgrksla, mbvcqryu$ differ in\nexactly two coordinates, so $qzxwvtnphjgrkslambvcqryu$ is an equilateral triangle, as\ndesired."
    },
    "kernel_variant": {
      "question": "Let $n\\ge 4$ be an integer.  Consider the $2^{n}$ vertices of the $n$–dimensional unit cube, i.e.\nall points of the form \\((x_{1},\\dots ,x_{n})\\) with each $x_{k}\\in\\{0,1\\}$.  \nSuppose $B$ is a subset of these vertices with\n\\[|B|\\;>\\;\\frac{3\\,2^{n}}{n}.\\]\nProve that $B$ contains three distinct points that are the vertices of an equilateral triangle.",
      "solution": "For every point P∈B define\nS_P = {cube vertices that differ from P in exactly one coordinate}.\n\n1. Each set S_P has exactly n elements, one for each coordinate that can be toggled. Hence\n   ∑_{P∈B} |S_P| = n·|B| > n·(3·2^n / n) = 3·2^n.\n\n2. The entire cube has only 2^n vertices, so the total just computed exceeds twice that number:\n   3·2^n > 2·2^n = 2^{n+1}.\n   By the pigeonhole principle, some vertex V must belong to at least three distinct sets, say\n   V ∈ S_P ∩ S_Q ∩ S_R with P, Q, R ∈ B all distinct.\n\n3. The vertex V differs from each of P, Q, R in exactly one coordinate.  Because P≠Q, the coordinates in which P and Q differ from V must be different; consequently P and Q differ in exactly two coordinates.  The same reasoning applies to the pairs (Q,R) and (R,P).\n\n4. In the unit cube, two vertices that differ in exactly two coordinates are at distance √2 apart.  Therefore\n   |PQ| = |QR| = |RP| = √2,\n   so the triangle PQR is equilateral.\n\nThus B indeed contains three vertices that form an equilateral triangle.",
      "_meta": {
        "core_steps": [
          "For every P in B, form the set S_P of the n cube-vertices that differ from P in exactly one coordinate.",
          "Count: Σ|S_P| = n·|B| > 2^{n+1}.",
          "Since the whole cube has only 2^n vertices, the pigeonhole principle forces some vertex to lie in at least three different S_P sets.",
          "Those three P’s all differ pairwise in exactly two coordinates, giving equal Euclidean distances—hence an equilateral triangle."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Lower bound on |B| chosen so that n·|B| exceeds twice the number of cube vertices; any constant c>2 would work.",
            "original": "2^{n+1}/n"
          },
          "slot2": {
            "description": "Actual numerical values assigned to cube coordinates; any two opposite numbers (e.g. 0 and 1, or ±a) preserve the argument.",
            "original": "±1"
          },
          "slot3": {
            "description": "Minimal dimension explicitly stated in the problem; any requirement n ≥ 3 (or any stricter n ≥ k with k ≥ 3) leaves the proof unchanged.",
            "original": "n ≥ 3"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}