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HMMT 二月 2006 · 冲刺赛 · 第 9 题

HMMT February 2006 — Guts Round — Problem 9

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

  1. [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?
    IX HARVARD-MIT MATHEMATICS TOURNAMENT, 25 FEBRUARY 2006 — GUTS ROUNDth 2

英文原题

[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?
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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IX th HARVARD-MIT MATHEMATICS TOURNAMENT, 25 FEBRUARY 2006 — GUTS ROUND

解析

英文解析

  1. 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?
    √
    Answer: + 1 − 3π
    Solution: The desired region consists of a small square and four “circle segments,”3
    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 segmentsare 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 circleπ 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