HMMT 二月 2016 · 团队赛 · 第 10 题
HMMT February 2016 — Team Round — Problem 10
题目详情
- [ 50 ] Let ABC be a triangle with incenter I whose incircle is tangent to BC , CA , AB at D , E , F . Point P lies on EF such that DP ⊥ EF . Ray BP meets AC at Y and ray CP meets AB at Z . Point Q is selected on the circumcircle of 4 AY Z so that AQ ⊥ BC . Prove that P , I , Q are collinear.
解析
- [ 50 ] Let ABC be a triangle with incenter I whose incircle is tangent to BC , CA , AB at D , E , F . Point P lies on EF such that DP ⊥ EF . Ray BP meets AC at Y and ray CP meets AB at Z . Point Q is selected on the circumcircle of 4 AY Z so that AQ ⊥ BC . Prove that P , I , Q are collinear. Proposed by: Evan Chen A X Y E Q Z P F I B C D The proof proceeds through a series of seven lemmas. Lemma 1. Lines DP and EF are the internal and external angle bisectors of ∠ BP C . Proof. Since DEF the cevian triangle of ABC with respect to its Gregonne point, we have that ( ) − 1 = EF ∩ BC, D ; B, C . ◦ Then since ∠ DP F = 90 we see P is on the Apollonian circle of BC through D . So the conclusion follows. Lemma 2. Triangles BP F and CEP are similar. Proof. Invoking the angle bisector theorem with the previous lemma gives BP BP CP CP = = = . BF BD CD CE But ∠ BF P = ∠ CEP , so 4 BF P ∼ 4 CEP . Lemma 3. Quadrilateral BZY C is cyclic; in particular, line Y Z is the antiparallel of line BC through ∠ BAC . Proof. Remark that ∠ Y BZ = ∠ P BF = ∠ ECP = ∠ Y CZ . Lemma 4. The circumcircles of triangles AY Z , AEF , ABC are concurrent at a point X such that 4 XBF ∼ 4 XCE . Proof. Note that line EF is the angle bisector of ∠ BP Z = ∠ CP Y . Thus ZF ZP Y P Y E = = = . F B P B P C EC Then, if we let X be the Miquel point of quadrilateral ZY CB , it follows that the spiral similarity mapping segment BZ to segment CY maps E to F ; therefore the circumcircle of 4 AEF must pass through X too. Lemma 5. Ray XP bisects ∠ F XE . Proof. The assertion amounts to XF BF F P = = . XE EC P E The first equality follows from the spiral similarity 4 BF X ∼ 4 CEX , while the second is from 4 BF P ∼ 4 CEP . So the proof is complete by the converse of angle bisector theorem. Lemma 6. Points X , P , I are collinear. 1 1 1 Proof. On one hand, ∠ F XI = ∠ F AI = ∠ A . On the other hand, ∠ F XP = ∠ F XE = ∠ A . Hence, 2 2 2 X , Y , I collinear. Lemma 7. Points X , Q , I are collinear. ◦ Proof. On one hand, ∠ AXQ = 90 , because we established earlier that line Y Z was antiparallel to ◦ line BC through ∠ A , hence AQ ⊥ BC means exactly that ∠ AZQ = AY Q = 90 . On the other hand, ◦ ∠ AXI = 90 according to the fact that X lies on the circle with diameter AI . This completes the proof of the lemma. Finally, combining the final two lemmas solves the problem.