HMMT 二月 2022 · 冲刺赛 · 第 12 题
HMMT February 2022 — Guts Round — Problem 12
题目详情
- [7] A unit square ABCD and a circle Γ have the following property: if P is a point in the plane not ◦ contained in the interior of Γ, then min( ∠ AP B, ∠ BP C, ∠ CP D, ∠ DP A ) ≤ 60 . The minimum possible aπ area of Γ can be expressed as for relatively prime positive integers a and b . Compute 100 a + b . b . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT February 2022, February 19, 2022 — GUTS ROUND Organization Team Team ID# 4 3 2
解析
- [7] A unit square ABCD and a circle Γ have the following property: if P is a point in the plane not ◦ contained in the interior of Γ, then min( ∠ AP B, ∠ BP C, ∠ CP D, ∠ DP A ) ≤ 60 . The minimum possible aπ area of Γ can be expressed as for relatively prime positive integers a and b . Compute 100 a + b . b Proposed by: Daniel Zhu Answer: 106 Solution: Note that the condition for Γ in the problem is equivalent to the following condition: if ◦ min( ∠ AP B, ∠ BP C, ∠ CP D, ∠ DP A ) > 60 , then P is contained in the interior of Γ. Let X , X , X , 1 2 3 and X be the four points in ABCD such that ABX , BCX , CDX , and DAX are all equilat- 4 1 2 3 4 eral triangles. Now, let Ω , Ω , Ω , and Ω be the respective circumcircles of these triangles, and 1 2 3 4 let the centers of these circles be O , O , O , and O . Note that the set of points P such that 1 2 3 4 ◦ ∠ AP B, ∠ BP C, ∠ CP D, ∠ DP A > 60 is the intersection of Ω , Ω , Ω , and Ω . We want to find 1 2 3 4 the area of the minimum circle containing this intersection. ′ ′ ′ ′ Let Γ and Γ intersect at B and B . Define C , D and A similarly. It is not hard to see that the 1 2 ′ ′ ′ ′ ′ ′ ′ ◦ circumcircle of square A B C D is the desired circle. Now observe that ∠ AB D = ∠ AB D = 60 . ′ ′ ◦ ′ ′ ′ ′ Similarly, ∠ AD B = 60 , so AB D is equilateral. Its height is the distance from A to B D , which is √ 1 6 π 6 π √ , so its side length is . This is also the diameter of the desired circle, so its area is · = . 3 4 9 6 2 4 3 2