{
  "index": "1966-B-1",
  "type": "GEO",
  "tag": [
    "GEO",
    "ALG"
  ],
  "difficulty": "",
  "question": "B-1. Let a convex polygon \\( P \\) be contained in a square of side one. Show that the sum of the squares of the sides of \\( P \\) is less than or equal to 4.",
  "solution": "B-1 Let \\( P_{1}, P_{2}, \\cdots, P_{n} \\) be the vertices of \\( P \\). Let \\( P_{1}^{\\prime}, P_{2}^{\\prime}, \\cdots, P_{n}^{\\prime} \\) be the projections of \\( P_{1}, P_{2}, \\cdots, P_{n} \\) upon one of the sides of the squares, and let \\( P_{1}^{\\prime \\prime}, P_{2}^{\\prime \\prime}, \\cdots, P_{n}^{\\prime \\prime} \\) be the projections of \\( P_{1}, P_{2}, \\cdots, P_{n} \\) upon a side that is orthogonal to the previous one. Since \\( P \\) is convex, the first side will be covered at most twice by the segments \\( \\overline{P_{1}^{\\prime} P_{2}^{\\prime}}, \\cdots, \\overline{P_{n-1}^{\\prime} P_{n}^{\\prime}}, \\overline{P_{n}^{\\prime} P_{1}^{\\prime}} \\). We thus deduce the inequality \\( \\overline{P_{1}^{\\prime} P_{2}^{\\prime 2}}+\\cdots+\\overline{P_{n}^{\\prime} P_{1}^{\\prime 2}} \\leqq 2 \\). Similarly \\( \\overline{P_{1}^{\\prime \\prime} P_{2}^{\\prime \\prime 2}}+\\cdots+\\overline{P_{n}^{\\prime \\prime} P_{1}^{\\prime \\prime}}{ }^{2} \\) \\( \\leqq 2 \\). Adding these two inequalities and using the Pythagorean theorem the assertion follows.",
  "vars": [
    "P",
    "P_1",
    "P_2",
    "P_n",
    "n"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "P": "polygon",
        "P_1": "vertexone",
        "P_2": "vertextwo",
        "P_n": "vertexn",
        "n": "sidecount"
      },
      "question": "B-1. Let a convex polygon \\( polygon \\) be contained in a square of side one. Show that the sum of the squares of the sides of \\( polygon \\) is less than or equal to 4.",
      "solution": "B-1 Let \\( vertexone, vertextwo, \\cdots, vertexn \\) be the vertices of \\( polygon \\). Let \\( vertexone^{\\prime}, vertextwo^{\\prime}, \\cdots, vertexn^{\\prime} \\) be the projections of \\( vertexone, vertextwo, \\cdots, vertexn \\) upon one of the sides of the squares, and let \\( vertexone^{\\prime \\prime}, vertextwo^{\\prime \\prime}, \\cdots, vertexn^{\\prime \\prime} \\) be the projections of \\( vertexone, vertextwo, \\cdots, vertexn \\) upon a side that is orthogonal to the previous one. Since \\( polygon \\) is convex, the first side will be covered at most twice by the segments \\( \\overline{vertexone^{\\prime} vertextwo^{\\prime}}, \\cdots, \\overline{polygon_{sidecount-1}^{\\prime} vertexn^{\\prime}}, \\overline{vertexn^{\\prime} vertexone^{\\prime}} \\). We thus deduce the inequality \\( \\overline{vertexone^{\\prime} vertextwo^{\\prime 2}}+\\cdots+\\overline{vertexn^{\\prime} vertexone^{\\prime 2}} \\leqq 2 \\). Similarly \\( \\overline{vertexone^{\\prime \\prime} vertextwo^{\\prime \\prime 2}}+\\cdots+\\overline{vertexn^{\\prime \\prime} vertexone^{\\prime \\prime}}{}^{2} \\leqq 2 \\). Adding these two inequalities and using the Pythagorean theorem the assertion follows."
    },
    "descriptive_long_confusing": {
      "map": {
        "P": "lighthouse",
        "P_1": "violinist",
        "P_2": "harmonica",
        "P_n": "cornfield",
        "n": "pendulum"
      },
      "question": "B-1. Let a convex polygon \\( lighthouse \\) be contained in a square of side one. Show that the sum of the squares of the sides of \\( lighthouse \\) is less than or equal to 4.",
      "solution": "B-1 Let \\( violinist, harmonica, \\cdots, cornfield \\) be the vertices of \\( lighthouse \\). Let \\( violinist^{\\prime}, harmonica^{\\prime}, \\cdots, cornfield^{\\prime} \\) be the projections of \\( violinist, harmonica, \\cdots, cornfield \\) upon one of the sides of the squares, and let \\( violinist^{\\prime \\prime}, harmonica^{\\prime \\prime}, \\cdots, cornfield^{\\prime \\prime} \\) be the projections of \\( violinist, harmonica, \\cdots, cornfield \\) upon a side that is orthogonal to the previous one. Since \\( lighthouse \\) is convex, the first side will be covered at most twice by the segments \\( \\overline{violinist^{\\prime} harmonica^{\\prime}}, \\cdots, \\overline{P_{pendulum-1}^{\\prime} cornfield^{\\prime}}, \\overline{cornfield^{\\prime} violinist^{\\prime}} \\). We thus deduce the inequality \\( \\overline{violinist^{\\prime} harmonica^{\\prime 2}}+\\cdots+\\overline{cornfield^{\\prime} violinist^{\\prime 2}} \\leqq 2 \\). Similarly \\( \\overline{violinist^{\\prime \\prime} harmonica^{\\prime \\prime 2}}+\\cdots+\\overline{cornfield^{\\prime \\prime} violinist^{\\prime \\prime}}{ }^{2} \\leqq 2 \\). Adding these two inequalities and using the Pythagorean theorem the assertion follows."
    },
    "descriptive_long_misleading": {
      "map": {
        "P": "amorphousshape",
        "P_1": "voidpointone",
        "P_2": "voidpointtwo",
        "P_n": "voidpointend",
        "n": "singular"
      },
      "question": "B-1. Let a convex polygon \\( amorphousshape \\) be contained in a square of side one. Show that the sum of the squares of the sides of \\( amorphousshape \\) is less than or equal to 4.",
      "solution": "B-1 Let \\( voidpointone, voidpointtwo, \\cdots, voidpointend \\) be the vertices of \\( amorphousshape \\). Let \\( voidpointone^{\\prime}, voidpointtwo^{\\prime}, \\cdots, voidpointend^{\\prime} \\) be the projections of \\( voidpointone, voidpointtwo, \\cdots, voidpointend \\) upon one of the sides of the squares, and let \\( voidpointone^{\\prime\\prime}, voidpointtwo^{\\prime\\prime}, \\cdots, voidpointend^{\\prime\\prime} \\) be the projections of \\( voidpointone, voidpointtwo, \\cdots, voidpointend \\) upon a side that is orthogonal to the previous one. Since \\( amorphousshape \\) is convex, the first side will be covered at most twice by the segments \\( \\overline{voidpointone^{\\prime} voidpointtwo^{\\prime}}, \\cdots, \\overline{P_{singular-1}^{\\prime} voidpointend^{\\prime}}, \\overline{voidpointend^{\\prime} voidpointone^{\\prime}} \\). We thus deduce the inequality \\( \\overline{voidpointone^{\\prime} voidpointtwo^{\\prime 2}}+\\cdots+\\overline{voidpointend^{\\prime} voidpointone^{\\prime 2}} \\leqq 2 \\). Similarly \\( \\overline{voidpointone^{\\prime\\prime} voidpointtwo^{\\prime\\prime 2}}+\\cdots+\\overline{voidpointend^{\\prime\\prime} voidpointone^{\\prime\\prime}}{ }^{2} \\) \\( \\leqq 2 \\). Adding these two inequalities and using the Pythagorean theorem the assertion follows."
    },
    "garbled_string": {
      "map": {
        "P": "pwrxqule",
        "P_1": "qabgtnmh",
        "P_2": "vhklesod",
        "P_n": "nzqtkfwa",
        "n": "mrevlusp"
      },
      "question": "B-1. Let a convex polygon \\( pwrxqule \\) be contained in a square of side one. Show that the sum of the squares of the sides of \\( pwrxqule \\) is less than or equal to 4.",
      "solution": "B-1 Let \\( qabgtnmh, vhklesod, \\cdots, nzqtkfwa \\) be the vertices of \\( pwrxqule \\). Let \\( qabgtnmh^{\\prime}, vhklesod^{\\prime}, \\cdots, nzqtkfwa^{\\prime} \\) be the projections of \\( qabgtnmh, vhklesod, \\cdots, nzqtkfwa \\) upon one of the sides of the squares, and let \\( qabgtnmh^{\\prime \\prime}, vhklesod^{\\prime \\prime}, \\cdots, nzqtkfwa^{\\prime \\prime} \\) be the projections of \\( qabgtnmh, vhklesod, \\cdots, nzqtkfwa \\) upon a side that is orthogonal to the previous one. Since \\( pwrxqule \\) is convex, the first side will be covered at most twice by the segments \\( \\overline{qabgtnmh^{\\prime} vhklesod^{\\prime}}, \\cdots, \\overline{pwrxqule_{mrevlusp-1}^{\\prime} nzqtkfwa^{\\prime}}, \\overline{nzqtkfwa^{\\prime} qabgtnmh^{\\prime}} \\). We thus deduce the inequality \\( \\overline{qabgtnmh^{\\prime} vhklesod^{\\prime 2}}+\\cdots+\\overline{nzqtkfwa^{\\prime} qabgtnmh^{\\prime 2}} \\leqq 2 \\). Similarly \\( \\overline{qabgtnmh^{\\prime \\prime} vhklesod^{\\prime \\prime 2}}+\\cdots+\\overline{nzqtkfwa^{\\prime \\prime} qabgtnmh^{\\prime \\prime}}{ }^{2} \\leqq 2 \\). Adding these two inequalities and using the Pythagorean theorem the assertion follows."
    },
    "kernel_variant": {
      "question": "Let \\(P\\) be a convex polygon that lies entirely inside an axis–aligned rectangle whose side–lengths are \\(2\\) (vertical) and \\(3\\) (horizontal). Prove that\n\\[\n\\sum_{\\text{sides }e\\text{ of }P} |e|^{2}\\;\\le\\;26.\n\\]",
      "solution": "Label the vertices of P cyclically by P₁,P₂,…,Pₙ.  Denote by Pᵢ′ the orthogonal projection of Pᵢ onto the bottom side of the rectangle (the 3–unit side), and by Pᵢ″ its projection onto the left side (the 2–unit side).\n\n1.  Projection onto the 3–unit side.\nBecause P is convex, the chain of projected segments\n   P₁′P₂′, P₂′P₃′, …, Pₙ′P₁′\ncan cover any point of that side at most twice.  Hence the total length of those projected segments is at most twice the side’s length,\n   ∑_{i=1}^n |Pᵢ′P_{i+1}′| ≤ 2·3 = 6.\nSince each |Pᵢ′P_{i+1}′| ≤ 3, it follows that\n   ∑_{i=1}^n |Pᵢ′P_{i+1}′|² ≤ (max |Pᵢ′P_{i+1}′|)·∑_{i=1}^n |Pᵢ′P_{i+1}′| ≤ 3·6 = 18.   (1)\n\n2.  Projection onto the 2–unit side.\nSimilarly, projecting onto the left side yields\n   ∑_{i=1}^n |Pᵢ″P_{i+1}″| ≤ 2·2=4,\nand since each |Pᵢ″P_{i+1}″| ≤ 2,\n   ∑_{i=1}^n |Pᵢ″P_{i+1}″|² ≤ 2·4=8.   (2)\n\n3.  Reconstructing the true edge lengths.\nFor each edge PᵢP_{i+1}, by the Pythagorean theorem,\n   |PᵢP_{i+1}|² = |Pᵢ′P_{i+1}′|² + |Pᵢ″P_{i+1}″|².\nSumming over all i and using (1) and (2) gives\n   ∑_{i=1}^n |PᵢP_{i+1}|² = ∑|Pᵢ′P_{i+1}′|² + ∑|Pᵢ″P_{i+1}″|² ≤ 18 + 8 = 26.\n\nTherefore the sum of the squares of the side–lengths of the convex polygon P does not exceed 26, as claimed.",
      "_meta": {
        "core_steps": [
          "Project every edge of the convex polygon onto one chosen axis (a side of the container).",
          "Use convexity to show each axis–side is covered at most twice, giving a bound for the sum of (squared) projection lengths in that direction.",
          "Repeat the same argument for the axis perpendicular to the first one.",
          "Apply the Pythagorean identity: edge-length² = (projection on axis 1)² + (projection on axis 2)².",
          "Add the two one-direction bounds to obtain the required global bound for the sum of the squared side-lengths."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Length of each side of the containing square (scales all bounds quadratically).",
            "original": "1"
          },
          "slot2": {
            "description": "Shape of the container, provided it supplies two orthogonal sides of lengths a and b (e.g., an axis-aligned rectangle instead of a square).",
            "original": "square"
          },
          "slot3": {
            "description": "Final numerical bound obtained for the sum of the squared side-lengths (depends on slot1 and slot2 values).",
            "original": "4"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}