HMMT 十一月 2018 · 冲刺赛 · 第 33 题

HMMT November 2018 — Guts Round — Problem 33

[ 17 ] Let ABC be a triangle with AB = 20 , BC = 10 , CA = 15. Let I be the incenter of ABC , and let BI meet AC at E and CI meet AB at F . Suppose that the circumcircles of BIF and CIE meet at a point D different from I . Find the length of the tangent from A to the circumcircle of DEF . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT November 2018, November 10, 2018 — GUTS ROUND Organization Team Team ID#

解析

[ 17 ] Let ABC be a triangle with AB = 20 , BC = 10 , CA = 15. Let I be the incenter of ABC , and let BI meet AC at E and CI meet AB at F . Suppose that the circumcircles of BIF and CIE meet at a point D di ↵ erent from I . Find the length of the tangent from A to the circumcircle of DEF . Proposed by: Michael Ren p Answer: 2 30 Solution 1: Let O = AI \ ( AEF ). We claim that O is the circumcenter of DEF . Indeed, note that \ B + \ C \ EOF \ EDF = \ ECI + \ F BI = = , and OE = OF , so the claim is proven. 2 2 Now note that the circumcircle of DEF passes through the incenter of AEF , so power of A with respect p to ( DEF ) is AE · AF . We can compute that AE = 10 , AF = 12 by the angle bisector theorem, so p p the length of the tangent is 10 · 12 = 2 30. 0 0 0 Solution 2: Let E be the reflection of E across AI . Note that E lies on AB . We claim that E lies on the circumcircle of DEF , which will imply that the power of A with respect to ( DEF ) is 0 AE · AF = AE · AF and we proceed as in Solution 1. We can easily compute \ A 0 \ EE F = 90 + 2 and \ A \ EDF = \ EDI + \ IDF = \ ECI + \ IBF = 90 . 2 0 0 Therefore \ EDF + \ EE F = 180 , so E lies on the circumcircle of DEF as desired.