PUMaC 2013 · 个人决赛（A 组） · 第 2 题

PUMaC 2013 — Individual Finals (Division A) — Problem 2

Let γ be the incircle of 4 ABC (i.e. the circle inscribed in 4 ABC ) and I be the center of γ . Let D , E and F be the feet of the perpendiculars from I to BC , CA and AB respectively. Let ′ ′ D be the point on γ such that DD is a diameter of γ . Suppose the tangent to γ through D ′ intersects the line EF at P . Suppose the tangent to γ through D intersects the line EF at ′ ◦ Q . Prove that ∠ P IQ + ∠ DAD = 180 .

解析

Let be the incircle of 4 ABC (i.e. the circle inscribed in 4 ABC ) and I be the center of . Let D , E and F be the feet of the perpendiculars from I to BC , CA and AB respectively. Let 0 0 D be the point on such that DD is a diameter of . Suppose the tangent to through D 0 intersects the line EF at P . Suppose the tangent to through D intersects the line EF at 0 Q . Prove that \ P IQ + \ DAD = 180 . 1 Solution 2 2 2 0 2 Let M be the point of intersection of A I and EF . Since IM · IA = IE = IF = ID = ID , it 0 0 0 0 follows that 4 IDM ⇠ 4 IAD and 4 ID M ⇠ 4 IAD , and hence \ DAD = \ DAI + \ D AI = 0 0 \ IDM + \ ID M . Since DIM P and D QIM are cyclic quadrilaterals, we have \ IDM = 0 0 \ IP M and \ ID M = \ IQM . Therefore, \ DAD = \ IP M + \ IQM = 180 \ P IQ as desired.