HMMT 二月 2004 · 冲刺赛 · 第 33 题

HMMT February 2004 — Guts Round — Problem 33

[10] A plane P slices through a cube of volume 1 with a cross-section in the shape of a regular hexagon. This cube also has an inscribed sphere, whose intersection with P is a circle. What is the area of the region inside the regular hexagon but outside the circle? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HARVARD-MIT MATHEMATICS TOURNAMENT, FEBRUARY 28, 2004 — GUTS ROUND

解析

A plane P slices through a cube of volume 1 with a cross-section in the shape of a regular hexagon. This cube also has an inscribed sphere, whose intersection with P is a circle. What is the area of the region inside the regular hexagon but outside the circle? √ Solution: (3 3 − π ) / 4 One can show that the hexagon must have as its vertices the midpoints of six edges of the cube, as illustrated; for example, this readily follows from the fact that opposite sides of the hexagons and the medians between them are parallel. We then conclude √ that the side of the hexagon is 2 / 2 (since it cuts off an isosceles triangle of leg 1 / 2 √ √ √ 2 from each face), so the area is (3 / 2)( 2 / 2) ( 3) = 3 3 / 4. Also, the plane passes through the center of the sphere by symmetry, so it cuts out a cross section of radius 1 / 2, whose area (which is contained entirely inside the hexagon) is then π/ 4. The √ sought area is thus (3 3 − π ) / 4.