HMMT 二月 2006 · 冲刺赛 · 第 28 题

HMMT February 2006 — Guts Round — Problem 28

[10] A pebble is shaped as the intersection of a cube of side length 1 with the solid sphere tangent to all of the cube’s edges. What is the surface area of this pebble? 2 2

解析

A pebble is shaped as the intersection of a cube of side length 1 with the solid sphere tangent to all of the cube’s edges. What is the surface area of this pebble? √ 6 2 − 5 Answer: π 2 Solution: Imagine drawing the sphere and the cube. Take a cross section, with a plane parallel to two of the cube’s faces, passing through the sphere’s center. In this cross section, the sphere looks like a circle, and the cube looks like a square (of side length 1) inscribed in that circle. We can now calculate that the sphere has diameter 9 √ 2 d := 2 and surface area S := πd = 2 π , and that the sphere protrudes a distance of √ 2 − 1 x := out from any given face of the cube. 2 It is known that the surface area chopped off from a sphere by any plane is proportional to the perpendicular distance thus chopped off. Thus, each face of the cube chops of x a fraction of the sphere’s surface. The surface area of the pebble contributed by the d x 1 sphere is thus S · (1 − 6 · ), whereas the cube contributes 6 circles of radius , with d 2 ( ) 2 1 3 total area 6 · π = π . The pebble’s surface area is therefore 2 2 ( ) √ √ ( ) x 3 2 − 1 3 6 2 − 5 S · 1 − 6 · + π = 2 π · 1 − 6 · √ + π = π. d 2 2 2 2 2 2 2