HMMT 二月 2016 · 冲刺赛 · 第 14 题

HMMT February 2016 — Guts Round — Problem 14

[ 9 ] Let ABC be a triangle such that AB = 13, BC = 14, CA = 15 and let E , F be the feet of the altitudes from B and C , respectively. Let the circumcircle of triangle AEF be ω . We draw three lines, tangent to the circumcircle of triangle AEF at A , E , and F . Compute the area of the triangle these three lines determine. ( ) ( ) ( ) π 2 π 3 π

解析

[ 9 ] Let ABC be a triangle such that AB = 13, BC = 14, CA = 15 and let E , F be the feet of the altitudes from B and C , respectively. Let the circumcircle of triangle AEF be ω . We draw three lines, tangent to the circumcircle of triangle AEF at A , E , and F . Compute the area of the triangle these three lines determine. Proposed by: Christopher Shao 462 Answer: 5 Note that AEF ∼ ABC . Let the vertices of the triangle whose area we wish to compute be P, Q, R , opposite A, E, F respectively. Since H, O are isogonal conjugates, line AH passes through the circum- center of AEF , so QR ‖ BC . Let M be the midpoint of BC . We claim that M = P . This can be seen by angle chasing at E, F to find that ∠ P F B = ∠ ABC , ∠ P EC = ∠ ACB , and noting that M is the circumcenter of BF EC . So, the height from P to QR is the height from A to BC , and thus if K is the area of ABC , the area we QR want is K . BC BC tan B Heron’s formula gives K = 84, and similar triangles QAF, M BF and RAE, M CE give QA = , 2 tan A QR BC tan C tan B +tan C tan B tan C − 1 11 RA = , so that = = = , 2 tan A BC 2 tan A 2 10 462 since the height from A to BC is 12. So our answer is . 5 ( ) ( ) ( ) π 2 π 3 π