{
  "index": "2013-A-1",
  "type": "COMB",
  "tag": [
    "COMB",
    "GEO"
  ],
  "difficulty": "",
  "question": "Recall that a regular icosahedron is a convex polyhedron\nhaving 12 vertices and 20 faces;\nthe faces are congruent equilateral triangles.\nOn each face of a regular icosahedron is written a nonnegative integer\nsuch that the sum of all 20 integers is 39. Show that there are \ntwo faces that share a vertex and have the same integer written on them.",
  "solution": "Suppose otherwise. Then each vertex $v$ is a vertex for five faces, all of which have different labels, and so the sum of the labels of the five faces incident to $v$ is at least $0+1+2+3+4 = 10$. Adding this sum over all vertices $v$ gives $3 \\times 39 = 117$, since each face's label is counted three times. Since there are $12$ vertices, we conclude that $10 \\times 12 \\leq 117$, contradiction.\n\n\\noindent\n\\textbf{Remark:}\nOne can also obtain the desired result by showing that any collection of five faces must contain two faces that share a vertex; it then follows that each label can appear at most $4$ times, and so the sum of all labels is at least $4(0+1+2+3+4) = 40 > 39$, contradiction.",
  "vars": [
    "v"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "v": "vertex"
      },
      "question": "Recall that a regular icosahedron is a convex polyhedron\nhaving 12 vertices and 20 faces;\nthe faces are congruent equilateral triangles.\nOn each face of a regular icosahedron is written a nonnegative integer\nsuch that the sum of all 20 integers is 39. Show that there are \ntwo faces that share a vertex and have the same integer written on them.",
      "solution": "Suppose otherwise. Then each vertex $vertex$ is a vertex for five faces, all of which have different labels, and so the sum of the labels of the five faces incident to $vertex$ is at least $0+1+2+3+4 = 10$. Adding this sum over all vertices $vertex$ gives $3 \\times 39 = 117$, since each face's label is counted three times. Since there are $12$ vertices, we conclude that $10 \\times 12 \\leq 117$, contradiction.\n\n\\noindent\n\\textbf{Remark:}\nOne can also obtain the desired result by showing that any collection of five faces must contain two faces that share a vertex; it then follows that each label can appear at most $4$ times, and so the sum of all labels is at least $4(0+1+2+3+4) = 40 > 39$, contradiction."
    },
    "descriptive_long_confusing": {
      "map": {
        "v": "marigold"
      },
      "question": "Recall that a regular icosahedron is a convex polyhedron\nhaving 12 vertices and 20 faces;\nthe faces are congruent equilateral triangles.\nOn each face of a regular icosahedron is written a nonnegative integer\nsuch that the sum of all 20 integers is 39. Show that there are \ntwo faces that share a vertex and have the same integer written on them.",
      "solution": "Suppose otherwise. Then each vertex $marigold$ is a vertex for five faces, all of which have different labels, and so the sum of the labels of the five faces incident to $marigold$ is at least $0+1+2+3+4 = 10$. Adding this sum over all vertices $marigold$ gives $3 \\times 39 = 117$, since each face's label is counted three times. Since there are $12$ vertices, we conclude that $10 \\times 12 \\leq 117$, contradiction.\n\n\\noindent\n\\textbf{Remark:}\nOne can also obtain the desired result by showing that any collection of five faces must contain two faces that share a vertex; it then follows that each label can appear at most $4$ times, and so the sum of all labels is at least $4(0+1+2+3+4) = 40 > 39$, contradiction."
    },
    "descriptive_long_misleading": {
      "map": {
        "v": "centerpoint"
      },
      "question": "\nRecall that a regular icosahedron is a convex polyhedron\nhaving 12 vertices and 20 faces;\nthe faces are congruent equilateral triangles.\nOn each face of a regular icosahedron is written a nonnegative integer\nsuch that the sum of all 20 integers is 39. Show that there are \ntwo faces that share a vertex and have the same integer written on them.\n",
      "solution": "\nSuppose otherwise. Then each vertex $centerpoint$ is a vertex for five faces, all of which have different labels, and so the sum of the labels of the five faces incident to $centerpoint$ is at least $0+1+2+3+4 = 10$. Adding this sum over all vertices $centerpoint$ gives $3 \\times 39 = 117$, since each face's label is counted three times. Since there are $12$ vertices, we conclude that $10 \\times 12 \\leq 117$, contradiction.\n\n\\noindent\n\\textbf{Remark:}\nOne can also obtain the desired result by showing that any collection of five faces must contain two faces that share a vertex; it then follows that each label can appear at most $4$ times, and so the sum of all labels is at least $4(0+1+2+3+4) = 40 > 39$, contradiction.\n"
    },
    "garbled_string": {
      "map": {
        "v": "qzxwvtnp"
      },
      "question": "Recall that a regular icosahedron is a convex polyhedron\nhaving 12 vertices and 20 faces;\nthe faces are congruent equilateral triangles.\nOn each face of a regular icosahedron is written a nonnegative integer\nsuch that the sum of all 20 integers is 39. Show that there are \ntwo faces that share a vertex and have the same integer written on them.",
      "solution": "Suppose otherwise. Then each vertex $qzxwvtnp$ is a vertex for five faces, all of which have different labels, and so the sum of the labels of the five faces incident to $qzxwvtnp$ is at least $0+1+2+3+4 = 10$. Adding this sum over all vertices $qzxwvtnp$ gives $3 \\times 39 = 117$, since each face's label is counted three times. Since there are $12$ vertices, we conclude that $10 \\times 12 \\leq 117$, contradiction.\n\n\\noindent\n\\textbf{Remark:}\nOne can also obtain the desired result by showing that any collection of five faces must contain two faces that share a vertex; it then follows that each label can appear at most $4$ times, and so the sum of all labels is at least $4(0+1+2+3+4) = 40 > 39$, contradiction."
    },
    "kernel_variant": {
      "question": "Recall that a (regular) cube is a convex polyhedron having 8 vertices and 6 faces; the faces are congruent squares.  On each face of a cube is written a positive integer (i.e., an integer \\(\\ge 1\\)) such that the sum of the six integers is \\(11\\).  Show that there exist two faces that share a vertex and have the same integer written on them.",
      "solution": "Assume, for the sake of contradiction, that no two faces meeting at a vertex carry the same label.\n\n1. Each vertex of a cube is incident to k=3 faces, and by the assumption these three labels are distinct positive integers.  The smallest possible choice for three distinct positive integers is 1,2,3, whose sum is 1+2+3 = 6.  Hence the sum of the labels on the faces incident to any vertex is at least 6.\n\n2. The cube has V = 8 vertices, so adding these vertex–sums over all vertices gives a total of at least 8×6 = 48.\n\n3. Each face of the cube is a square, so it meets t = 4 vertices.  Consequently every face–label is counted exactly four times in the total found in step 2.  Denoting the total of all six labels by S, we therefore have the inequality\n   4S ≥ 48.\n\n4. But the problem states that the six labels sum to S = 11, for which 4S = 44 < 48, contradicting the inequality just obtained.\n\n5. The contradiction shows that our initial assumption was false.  Therefore two faces that share a vertex must indeed carry the same integer label.\n\nThus the statement is proved.",
      "_meta": {
        "core_steps": [
          "Assume, for contradiction, that no two faces sharing a vertex carry the same label.",
          "Because the k faces incident to any vertex must then have distinct non-negative labels, the sum at each vertex is at least 0+1+⋯+(k−1)=k(k−1)/2.",
          "Sum these vertex–sums over all V vertices; each face is counted exactly t times (t = #vertices per face). Hence t·(total label sum) ≥ V·k(k−1)/2.",
          "Insert the given upper bound on the total label sum; this yields an inequality that fails (contradiction).",
          "Therefore two faces sharing a vertex must have the same label."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "#faces meeting at each vertex (vertex degree k)",
            "original": 5
          },
          "slot2": {
            "description": "#vertices of each face (t, used in double-counting)",
            "original": 3
          },
          "slot3": {
            "description": "total number of vertices V of the polyhedron",
            "original": 12
          },
          "slot4": {
            "description": "stated upper bound on the sum of all labels",
            "original": 39
          },
          "slot5": {
            "description": "least admissible label (non-negativity ⇒ smallest possible label)",
            "original": 0
          },
          "slot6": {
            "description": "requirement that faces be congruent equilateral triangles (metric regularity, not used in proof)",
            "original": "faces are congruent equilateral triangles"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}