HMMT 十一月 2012 · 冲刺赛 · 第 12 题

HMMT November 2012 — Guts Round — Problem 12

[ 8 ] Three circles of radius 1 are drawn, whose centers form the vertices of an equilateral triangle of side length 1. Find the area of the region common to at least two of the circles. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT NOVEMBER 2012, 10 NOVEMBER 2012 — GUTS ROUND ◦

解析

[ 8 ] √ 2 π − 3 Answer: Let A be the equilateral triangle. Let A be the area of the region outside of △ 1 2 the equilateral triangle but inside the second and third circles. Define A , A analogously. We have 2 3 A = A = A = A = 1 2 3 k √ ( ) 2 2 1 · π 1 · sin 120 4 π − 3 3 − = , 3 2 12 and √ 2 1 · sin 60 3 A = = . △ 2 4 Thus, the total area is √ √ √ 4 π − 3 3 3 2 π − 3 A + A + A + A = 3 · + = . 1 2 3 △ 12 4 2