HMMT 二月 2021 · 冲刺赛 · 第 26 题

HMMT February 2021 — Guts Round — Problem 26

[16] Let triangle ABC have incircle ω , which touches BC , CA , and AB at D , E , and F , respectively. Then, let ω and ω be circles tangent to AD and internally tangent to ω at E and F , respectively. Let 1 2 P be the intersection of line EF and the line passing through the centers of ω and ω . If ω and ω have 1 2 1 2 radii 5 and 6, respectively, compute P E · P F .

解析

[16] Let triangle ABC have incircle ω , which touches BC , CA , and AB at D , E , and F , respectively. Then, let ω and ω be circles tangent to AD and internally tangent to ω at E and F , respectively. 1 2 Let P be the intersection of line EF and the line passing through the centers of ω and ω . If ω and 1 2 1 ω have radii 5 and 6, respectively, compute P E · P F . 2 Proposed by: Akash Das Answer: 3600 Solution: Let the centers of ω and ω be O and O . Let DE intersect ω again at Q , and let DF 1 2 1 2 1 intersect ω again at R . Note that since ω and ω must be tangent to AD at the same point (by equal 2 1 2 tangents), so AD must be the radical axis of ω and ω , so RQEF is cyclic. Thus, we have 1 2 ◦ ◦ ∠ O QR = ∠ EQR − ∠ O QE = 180 − ∠ EF D − ∠ O EQ = 90 1 1 1 Thus, we have QR is tangent to ω , and similarly it must be tangent to ω as well. 1 2 Now, note that by Monge’s theorem on ω , ω , and ω , we have that P must be the intersection of the 1 2 external tangents of ω and ω . Since RQ is an external tangent, we have P , Q , and R are collinear. 1 2 √ √ Thus, by power of a point, we have P E · P F = P R · P Q . Note that P R = 10 30 and P Q = 12 30. Thus, we have P E · P F = 3600.