HMMT 二月 2006 · 冲刺赛 · 第 36 题

英文解析

36. Four points are independently chosen uniformly at random from the interior of a regulardodecahedron. What is the probability that they form a tetrahedron whose interiorcontains the dodecahedron’s center?
    Answer:1
    Solution: Situate the origin O at the dodecahedron’s center, and call the four random 8
    points P , where 1 ≤ i ≤ 4.
    To any tetrahedron P P P P we can associate a quadruple ( ), where ( ijk ) rangesi
    1 2 3 4 ( ijk )
    over all conjugates of the cycle (123) in the alternating group A : is the sign of
    4 ijkthe directed volume [ OP P P ]. Assume that, for a given tetrahedron P P P P , alli 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 theitetrahedron’s quadruple that involve P all flip sign, because each [ OP P P ] is a lineari i j k − − →12
    function of the vector OP . Thus, if we consider the 16 sister tetrahedra obtainediby choosing independently whether to flip each P through the origin, the quadruplesirange through all 16 possibilities (namely, all the quadruples consisting of ± 1 s). Twoof 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.