HMMT 二月 2008 · 冲刺赛 · 第 28 题

英文解析

28. [ 15 ] Let P be a polyhedron where every face is a regular polygon, and every edge has length 1. Eachvertex of P is incident to two regular hexagons and one square. Choose a vertex V of the polyhedron.
    Find the volume of the set of all points contained in P that are closer to V than to any other vertex.
    √
    Answer: Observe that P is a truncated octahedron, formed by cutting off the corners from 2
    a regular octahedron with edge length 3. So, to compute the value of P , we can find the volume of 3
    the octahedron, and then subtract off the volume of truncated corners. Given a square pyramid whereeach triangular face an equilateral triangle, and whose side length is s , the height of the pyramid is
    √ √ √
    2 1 2 2
    2 3
    s , and thus the volume is · s · s = s . The side length of the octahedron is 3, and noting
    2 3 2 6
    √
    √
    2(3)3
    that the octahedron is made up of two square pyramids, its volume must be is 2 · = 9 2.
    √6
    The six “corners” that we remove are all square pyramids, each with volume , and so the resulting 2
    √6
    √ √
    polyhedron P has volume 9 2 − 6 · = 8 2.2
    Finally, to find the volume of all points closer to one particular vertex than any other vertex, note 6
    that due to symmetry, every point in P (except for a set with zero volume), is closest to one of the 24
    vertices. Due to symmetry, it doesn’t matter which V is picked, so we can just divide the volume of P
    √
    by 24 to obtain the answer .2 3