HMMT 二月 2008 · 冲刺赛 · 第 28 题

HMMT February 2008 — Guts Round — Problem 28

[ 15 ] Let P be a polyhedron where every face is a regular polygon, and every edge has length 1. Each vertex 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.

解析

[ 15 ] Let P be a polyhedron where every face is a regular polygon, and every edge has length 1. Each vertex 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. √ 2 Answer: Observe that P is a truncated octahedron, formed by cutting off the corners from 3 a regular octahedron with edge length 3. So, to compute the value of P , we can find the volume of the octahedron, and then subtract off the volume of truncated corners. Given a square pyramid where each 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 √ √ 3 2(3) that the octahedron is made up of two square pyramids, its volume must be is 2 · = 9 2. 6 √ 2 The six “corners” that we remove are all square pyramids, each with volume , and so the resulting 6 √ √ √ 2 polyhedron P has volume 9 2 − 6 · = 8 2. 6 Finally, to find the volume of all points closer to one particular vertex than any other vertex, note 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 √ 2 by 24 to obtain the answer . 3