HMMT 二月 2017 · 几何 · 第 5 题

HMMT February 2017 — Geometry — Problem 5

Let ABCD be a quadrilateral with an inscribed circle ω and let P be the intersection of its diagonals AC and BD . Let R , R , R , R be the circumradii of triangles AP B , BP C , CP D , DP A respectively. 1 2 3 4 If R = 31 and R = 24 and R = 12, find R . 1 2 3 4

解析

Let ABCD be a quadrilateral with an inscribed circle ω and let P be the intersection of its diagonals AC and BD . Let R , R , R , R be the circumradii of triangles AP B , BP C , CP D , DP A respectively. 1 2 3 4 If R = 31 and R = 24 and R = 12, find R . 1 2 3 4 Proposed by: Sam Korsky Answer: 19 ◦ ◦ Note that ∠ AP B = 180 − ∠ BP C = ∠ CP D = 180 − ∠ DP A so sin AP B = sin BP C = sin CP D = sin DP A . Now let ω touch sides AB, BC, CD, DA at E, F, G, H respectively. Then AB + CD = AE + BF + CG + DH = BC + DA so AB CD BC DA

= + sin AP B sin CP D sin BP C sin DP A and by the Extended Law of Sines this implies 2 R + 2 R = 2 R + 2 R 1 3 2 4 which immediately yields R = R + R − R = 19 . 4 1 3 2