PUMaC 2012 · 个人决赛（A 组） · 第 3 题
PUMaC 2012 — Individual Finals (Division A) — Problem 3
题目详情
- Let ABC be a triangle with incenter I , and let D be the foot of the angle bisector from A to BC . Let Γ be the circumcircle of triangle BIC , and let P Q be a chord of Γ passing through D . Prove that AD bisects ∠ P AQ . Please write complete, concise and clear proofs. Have fun! – PUMaC Problem Writers 1
解析
- Let ABC be a triangle with incenter I , and let D be the foot of the angle bisector from A to BC . Let Γ be the circumcircle of triangle BIC , and let P Q be a chord of Γ passing through D . Prove that AD bisects ∠ P AQ . 1 Solution: If P Q = BC the result is trivial, so we may assume otherwise. Let m ∠ ABC = b , m ∠ BCA = c , and m ∠ CAB = a . Also, let T be the center of Γ and let c Ω be the circumcircle of triangle ABC . We first claim that T lies on Ω. As m ∠ ICB = , 2 we have m ∠ IT B = c . Similarly, m ∠ IT C = b . Thus m ∠ BAC + m ∠ BT C = a + ( b + c ) = π , so quadrilateral ACT B is cyclic. As BT and CT are chords of Ω with equal length, we must ← − → have m ∠ BAT = m ∠ CAT , so T lies on line AID . We now wish to show that quadrilateral AQT P is cyclic. Let Λ be the circumcircle of triangle AP Q . Since one of P, Q lies inside Ω and the other lies outside Ω, Λ and Ω must intersect in ′ exactly two points, and we let the point of intersection which is not A be called T . As Λ and ′ Ω have radical axis AT , Λ and Γ have radical axis P Q , and Ω and Γ have radical axis BC , it follows by the radical axis theorem that these three line segments must be concurrent. As ′ ′ BC and P Q intersect at point D , we see that AT must pass through D , so T lies both on ← → ← → ′ ′ AD and on Ω. AD and Ω intersect only at A and T , and as T 6 = A , it follows that T = T . Thus AQT P is cyclic. ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ As P T = QT , we see that m ∠ T P Q = m ∠ T QP . Since AQT P is cyclic, we conclude that m ∠ DAP = m ∠ T AP = m ∠ T QP = m ∠ T P Q = m ∠ T AQ = m ∠ DAQ , and the result follows. 2