HMMT 十一月 2020 · 冲刺赛 · 第 20 题

HMMT November 2020 — Guts Round — Problem 20

[11] Let ω be a circle of radius 5, and let ω be a circle of radius 2 whose center lies on ω . Let the two 1 2 1 circles intersect at A and B , and let the tangents to ω at A and B intersect at P . If the area of 4 ABP 2 √ a b can be expressed as , where b is square-free and a, c are relatively prime positive integers, compute c 100 a + 10 b + c .

解析

[11] Let ω be a circle of radius 5, and let ω be a circle of radius 2 whose center lies on ω . Let the 1 2 1 two circles intersect at A and B , and let the tangents to ω at A and B intersect at P . If the area of 2 √ a b 4 ABP can be expressed as , where b is square-free and a, c are relatively prime positive integers, c compute 100 a + 10 b + c . Proposed by: Hahn Lheem Answer: 19285 Solution: A O P 2 O 1 B Let O and O be the centers of ω and ω , respectively. Because 1 2 1 2 ◦ ◦ ◦ ∠ O AP + ∠ O BP = 90 + 90 = 180 , 2 2 quadrilateral O AP B is cyclic. But O , A , and B lie on ω , so P lies on ω and O P is a diameter of 2 2 1 1 2 ω . 1 √ √ 2 6 From the Pythagorean theorem on triangle P AO , we can calculate AP = 4 6, so sin ∠ AO P = 2 2 5 1 and cos ∠ AO P = . Because 4 AO P and 4 BO P are congruent, we have 2 2 2 5 √ 4 6 sin ∠ AP B = sin 2 ∠ AO P = 2 sin ∠ AO P cos ∠ AO P = , 2 2 2 25 implying that √ P A · P B 192 6 [ AP B ] = sin ∠ AP B = . 2 25