HMMT 二月 2006 · 冲刺赛 · 第 9 题

HMMT February 2006 — Guts Round — Problem 9

[6] Four unit circles are centered at the vertices of a unit square, one circle at each vertex. What is the area of the region common to all four circles? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . th IX HARVARD-MIT MATHEMATICS TOURNAMENT, 25 FEBRUARY 2006 — GUTS ROUND 2

解析

Four unit circles are centered at the vertices of a unit square, one circle at each vertex. What is the area of the region common to all four circles? √ π Answer: + 1 − 3 3 Solution: The desired region consists of a small square and four “circle segments,” i.e. regions of a circle bounded by a chord and an arc. The side of this small square ◦ is just the chord of a unit circle that cuts off an angle of 30 , and the circle segments are bounded by that chord and the circle. Using the law of cosines (in an isosceles ◦ triangle with unit leg length and vertex angle 30 ), we find that the square of the √ length of the chord is equal to 2 − 3. We can also compute the area of each circle π 1 π 1 ◦ segment, namely − (1)(1) sin 30 = − . Hence, the desired region has area 12 2 12 4 ( ) √ √ π 1 π 2 − 3 + 4 − = + 1 − 3. 12 4 3 2