HMMT 十一月 2018 · 团队赛 · 第 6 题

HMMT November 2018 — Team Round — Problem 6

[ 45 ] Triangle 4 P QR , with P Q = P R = 5 and QR = 6, is inscribed in circle ω . Compute the radius of the circle with center on QR which is tangent to both ω and P Q .

解析

[ 45 ] Triangle 4 P QR , with P Q = P R = 5 and QR = 6, is inscribed in circle ω . Compute the radius of the circle with center on QR which is tangent to both ω and P Q . Proposed by: Michael Tang 20 Answer: 9 Solution 1: Denote the second circle by γ . Let T and r be the center and radius of γ , respectively, and let X and H be the tangency points of γ with ω and P Q , respectively. Let O be the center of ω , and 1 let M be the midpoint of QR . Note that QM = M R = QR = 3, so 4 P M Q and 4 P M R are 3 − 4 − 5 2 5 5 triangles. Since 4 QHT ∼ 4 QM P and HT = r , we get QT = r . Then T M = QM − QT = 3 − r . 4 4 P R 5 25 By the extended law of sines, the circumradius of 4 P QR is OP = = = , so 2 sin ∠ P QR 2(4 / 5) 8 25 7 25 OM = M P − OP = 4 − = . Also, we have OT = OX − XT = − r . Therefore, by the 8 8 8 Pythagorean theorem, ( ) ( ) ( ) 2 2 2 5 7 25 3 − r + = − r . 4 8 8 9 5 5 16 20 2 This simplifies to r − r = 0, so r = · = . 16 4 4 9 9 P H O Q T M R X Solution 2: Following the notation of the previous solution, we compute the power of T with respect to ( ) 5 5 ω in two ways. One of these is − QT · T R = − r 6 − r . The other way is − ( OX − OT )( OX + OT ) = 4 4 25 20 − r ( − r ). Equating these two yields r = . 4 9