HMMT 二月 2003 · 几何 · 第 8 题

HMMT February 2003 — Geometry — Problem 8

Let ABC be an equilateral triangle of side length 2. Let ω be its circumcircle, and let ω , ω , ω be circles congruent to ω centered at each of its vertices. Let R be the set A B C of all points in the plane contained in exactly two of these four circles. What is the area of R ? ◦ ◦ 6 6

解析

Let ABC be an equilateral triangle of side length 2. Let ω be its circumcircle, and let ω , ω , ω be circles congruent to ω centered at each of its vertices. Let R be the set A B C of all points in the plane contained in exactly two of these four circles. What is the area of R ? √ Solution: 2 3 ω , ω , ω intersect at the circumcenter; thus, every point within the circumcircle, and A B C no point outside of it, is in two or more circles. The area inside exactly two circles is shaded in the figure. The two intersection points of ω and ω , together with A , A B form the vertices of an equilateral triangle. As shown, this equilateral triangle cuts off ◦ a “lip” of ω (bounded by a 60 arc of ω and the corresponding chord) and another, congruent lip of ω that is not part of the region of interest. By rotating the first lip B to the position of the second, we can reassemble the equilateral triangle. Doing this for each of the 6 such triangles, we see that the desired area equals the area of a regular √ √ hexagon inscribed in ω . The side length of this hexagon is (2 / 3) · ( 3 / 2) · 2 = 2 3 / 3, √ √ √ 2 so its area is 6 · ( 3 / 4) · (2 3 / 3) = 2 3, and this is the answer. 3 ω A A ω ω B C B C ω ◦ ◦ 6 6