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HMMT 二月 2006 · 冲刺赛 · 第 36 题

HMMT February 2006 — Guts Round — Problem 36

专题
Discrete Math / 离散数学
难度
L3
来源
HMMT

题目详情

  1. [12] Four points are independently chosen uniformly at random from the interior of a regular dodecahedron. What is the probability that they form a tetrahedron whose interior contains the dodecahedron’s center? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . th IX HARVARD-MIT MATHEMATICS TOURNAMENT, 25 FEBRUARY 2006 — GUTS ROUND ∞ ∑ 2 n + 5
解析
  1. Four points are independently chosen uniformly at random from the interior of a regular dodecahedron. What is the probability that they form a tetrahedron whose interior contains the dodecahedron’s center? 1 Answer: 8 Solution: Situate the origin O at the dodecahedron’s center, and call the four random points P , where 1 ≤ i ≤ 4. i To any tetrahedron P P P P we can associate a quadruple ( ), where ( ijk ) ranges 1 2 3 4 ( ijk ) over all conjugates of the cycle (123) in the alternating group A : is the sign of 4 ijk the directed volume [ OP P P ]. Assume that, for a given tetrahedron P P P P , all i j k 1 2 3 4 members of its quadruple are nonzero (this happens with probability 1). For 1 ≤ i ≤ 4, if we replace P with its reflection through the origin, the three members of the i tetrahedron’s quadruple that involve P all flip sign, because each [ OP P P ] is a linear i i j k 12 − − → function of the vector OP . Thus, if we consider the 16 sister tetrahedra obtained i by choosing independently whether to flip each P through the origin, the quadruples i range through all 16 possibilities (namely, all the quadruples consisting of ± 1s). Two of these 16 tetrahedra, namely those with quadruples (1 , 1 , 1 , 1) and ( − 1 , − 1 , − 1 , − 1), will contain the origin. So the answer is 2 / 16 = 1 / 8.