HMMT 二月 2025 · 团队赛 · 第 8 题
HMMT February 2025 — Team Round — Problem 8
题目详情
- [50] Let △ ABC be a triangle with incenter I . The incircle of triangle △ ABC touches BC at D . Let M be the midpoint of BC , and let line AI meet the circumcircle of triangle △ ABC again at L ̸ = A . Let ω be the circle centered at L tangent to AB and AC . If ω intersects segment AD at point P , prove ◦ that ∠ IP M = 90 .
解析
- [50] Let △ ABC be a triangle with incenter I . The incircle of triangle △ ABC touches BC at D . Let M be the midpoint of BC , and let line AI meet the circumcircle of triangle △ ABC again at L ̸ = A . Let ω be the circle centered at L tangent to AB and AC . If ω intersects segment AD at point P , prove ◦ that ∠ IP M = 90 . Proposed by: Pitchayut Saengrungkongka Solution 1: Let X and Y be the bottom and top point on ω (i.e., the tangents of X and Y to ω are parallel to BC , and Y and A lie on the same side of BC ). Note that A , P , D , and X are collinear by homothety between the incircle and ω . The key claim is the following. Claim 1. Line IY is tangent to ω . ′ ′ Proof. Let the line through I parallel to BC meet AB and AC at B and C , respectively. Notice that ′ ′ ′ ′ ′ B L is the perpendicular bisector of BI , so B L externally bisects ∠ AB C . Similarly, C L externally ′ ′ ′ ′ ′ ′ bisects ∠ AC B . Hence, L is the excenter of △ AB C , which means that B C is tangent to ω . A ′ Y I C ′ B P C B D M L X Now, we note that LY ⊥ BC , so L , Y , and M are collinear (on the perpendicular bisector of BC ). ◦ ◦ Since ∠ Y P X = 90 and ∠ Y M D = 90 , P DM Y is cyclic. However, IY M D is a rectangle, so IP DM Y ◦ is a cyclic pentagon. Hence, ∠ IP M = ∠ IDM = 90 . Solution 2: Let the incircle touch AC and AB at E and F , respectively. Let DI intersect EF at X . ′ Let D be the other intersection of AD and the incircle. ′ Claim 2. P M ∥ D X . ′ Proof. Consider the homothety at A that sends ω to the incircle. It sends L to I and P to D . Furthermore, it’s well-known that X lies on AM . Because IX ∥ LM , we also have that the homothety ′ sends M to X . These facts imply that D X ∥ P M . A T ′ D X I P ′ T B C M D L ′ Let T be the antipode of D on the incircle. Let AT intersect the incircle again at T . Since X lies on ′ ′ the polar of A with respect to the incircle, by Brocard’s theorem, we have D , X , and T are collinear. ′ ′ ′ ′ It is well-known that AT ∥ IM . Therefore, ∠ DP M = ∠ DD X = ∠ DD T = ∠ DT T = ∠ DIM. ◦ Consequently, IM DP is cyclic, and ∠ IP M = ∠ IDM = 90 . Solution 3: Let ω be tangent to AB and AC at E and F , respectively. Note that these are the feet of the altitudes from L to AB and AC , and L lies on the circumcircle of △ ABC by Fact 5. As M is clearly the foot from L to BC , it follows that E , F , and M are collinear on the Simson Line of L with respect to △ ABC . Lastly, we want P to be on the circle with diameter IM . This circle intersects EF again at the foot ′ from M to AI , which is the midpoint of EF . Let this point be M . Consider the homothety sending ′ the incircle to ω . This clearly sends D to the second intersection of AD and ω , which is P , and it ′ 2 sends I to L . Note that AP · AP = AE = AM · AL , as the circle with diameter LE is tangent to ′ ′ ′ ′ ′ AE . Thus, P P M L is cyclic. Since ID ∥ LP , I lies on M L , and D lies on P P . By Reim’s, we also ′ ′ ◦ have P DM I is cyclic. As IM is a diameter of ( DM I ), we have ∠ IP M = 90 . A I P F D C B M ′ M E L ′ P