HMMT 二月 2003 · 几何 · 第 8 题

英文解析

8. Let ABC be an equilateral triangle of side length 2. Let ω be its circumcircle, and letω , ω , ω be circles congruent to ω centered at each of its vertices. Let R be the set
   A B Cof 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 Cno point outside of it, is in two or more circles. The area inside exactly two circlesis shaded in the figure. The two intersection points of ω and ω , together with A ,
   A Bform 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 lipto the position of the second, we can reassemble the equilateral triangle. Doing this for beach 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,
   √ √ √
   so its area is 6 · ( 3 / 4) · (2 3 / 3) = 2 3, and this is the answer.2
   ω3
   ωωAA
   CBCBω
   ° °
   6 6