HMMT 二月 2005 · 冲刺赛 · 第 30 题

HMMT February 2005 — Guts Round — Problem 30

[10] A cuboctahedron is a polyhedron whose faces are squares and equilateral triangles such that two squares and two triangles alternate around each vertex, as shown. What is the volume of a cuboctahedron of side length 1? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HARVARD-MIT MATHEMATICS TOURNAMENT, FEBRUARY 19, 2005 — GUTS ROUND

解析

A cuboctahedron is a polyhedron whose faces are squares and equilateral triangles such that two squares and two triangles alternate around each vertex, as shown. What is the volume of a cuboctahedron of side length 1? √ Solution: 5 2 / 3 We can construct a cube such that the vertices of the cuboctahedron are the midpoints of the edges of the cube. Let s be the side length of this cube. Now, the cuboctahedron is obtained from the cube by cutting a tetrahedron from each corner. Each such tetrahedron has a base in 2 the form of an isosceles right triangle of area ( s/ 2) / 2 and height s/ 2 for a volume of 3 ( s/ 2) / 6. The total volume of the cuboctahedron is therefore 3 3 3 s − 8 · ( s/ 2) / 6 = 5 s / 6 . Now, the side of the cuboctahedron is the hypotenuse of an isosceles right triangle √ √ of leg s/ 2; thus 1 = ( s/ 2) 2, giving s = 2, so the volume of the cuboctahedron is √ 5 2 / 3. 11