HMMT 二月 2023 · 团队赛 · 第 7 题
HMMT February 2023 — Team Round — Problem 7
题目详情
- [55] Let ABC be a triangle. Point D lies on segment BC such that ∠ BAD = ∠ DAC . Point X lies on the opposite side of line BC as A and satisfies XB = XD and ∠ BXD = ∠ ACB . Analogously, point Y lies on the opposite side of line BC as A and satisfies Y C = Y D and ∠ CY D = ∠ ABC . Prove that lines XY and AD are perpendicular.
解析
- [55] Let ABC be a triangle. Point D lies on segment BC such that ∠ BAD = ∠ DAC . Point X lies on the opposite side of line BC as A and satisfies XB = XD and ∠ BXD = ∠ ACB . Analogously, point Y lies on the opposite side of line BC as A and satisfies Y C = Y D and ∠ CY D = ∠ ABC . Prove that lines XY and AD are perpendicular. Proposed by: Pitchayut Saengrungkongka Solution 1: A D B C Y X I A Let I and I be the incenter and the A -excenter of 4 ABC . The key observation is that X is the A circumcenter of 4 BDI . To see why this is true, note that A ∠ BXD = ∠ C = 2 ∠ ICB = 2 ∠ II B = 2 ∠ DI B. A A Analogously, Y is the circumcenter of 4 CDI . Hence, XY is the perpendicular bisector of DI , which A A is clearly perpendicular to AD . Solution 2: A D B C M Y X Denote ω and ω as the circumcircle of 4 BXD and 4 CY D . Also, let AD intersects the circumcircle B C of 4 ABC at M . Since ∠ BXD = ∠ ACB = ∠ AM B , we get that M ∈ ω . Similarly, M ∈ ω . From B C here, there are two ways to finish. • Note by Law of Sine that the radius of ω and ω are DB/ (2 sin ∠ M DB ) and DB/ (2 sin ∠ M DC ), B C so they are actually equal. Thus, if O and O are the centers of ω and ω , then XO = Y O . B C B C B C Moreover, XO and Y O are both clearly perpendicular to BC , so XO O Y is parallelogram, B C B C implying that XY ‖ O O ⊥ DM . B C • Observe that ∠ A ◦ ∠ M XD = ∠ M BD = = 90 − ∠ XDY, 2 so XM ⊥ DY . Similarly, Y M ⊥ XD , so M is the orthocenter of 4 DXY , implying the result. Solution 3: A Q P I D B C Y X Let DX intersects AC at P and DY intersects AB at Q . Observe that ∠ BCP = ∠ BXP , so B, C, P, X are concyclic. This implies that CD = CP and that DB · DC = DX · DP . Similarly, we have BD = BQ and that DB · DC = DY · DQ . Thus, we actually have DX · DP = DY · DQ , implying that X, Y, P, Q are concyclic. Now, let I be the incenter of 4 ABC . Since BD = BQ , it follows that BI is the perpendicular bisector of DQ , so ID = IQ . Similarly, ID = IP , so I is actually the circumcenter of 4 DP Q . We then finish by angle chasing: ◦ ◦ ◦ ∠ XDI = 180 − ∠ P DI = 90 + ∠ DQP = 90 + ∠ DXY, implying the result.