HMMT 二月 2026 · 几何 · 第 8 题
HMMT February 2026 — Geometry — Problem 8
题目详情
- Let ABC be a triangle with orthocenter H . The internal angle bisector of ∠ BAC meets the circum- circles of triangles ABH , ACH , and ABC again at points P , Q , and M , respectively. Suppose that points A , P , Q , and M are distinct and lie on the internal angle bisector of ∠ BAC in that order. Given that AP = 4 , AQ = 5 , and BC = 7 , compute AM .
解析
- Let ABC be a triangle with orthocenter H . The internal angle bisector of ∠ BAC meets the circum- circles of triangles ABH , ACH , and ABC again at points P , Q , and M , respectively. Suppose that points A , P , Q , and M are distinct and lie on the internal angle bisector of ∠ BAC in that order. Given that AP = 4 , AQ = 5 , and BC = 7 , compute AM . Proposed by: Pitchayut Saengrungkongka √ Answer: 69 Solution 1: A ′ Q ′ P P H Q C B M Because H is the orthocenter of △ ABC , the circumcircles of △ ABH is the reflection of the circumcircle ′ of △ ABC across the midpoint of AB . Let P be the reflection of P across the midpoint of AB , then ′ ′ P must lie on the circumcircle of △ ABC . Similarly, let Q be the reflection of Q across the midpoint ′ of AC , then Q must lie on the circumcircle of △ ABC too. ′ ′ ′ ′ ′ ′ Note that BP = 4 , CQ = 5 . Moreover, P B ∥ AM ∥ Q C , so P Q = BC = 7 . Furthermore, since ′ ′ ′ ∠ BCQ = ∠ ACM , we have AM = BQ . Similarly, we also have AM = CP . ′ ′ ′ ′ ′ ′ ′ ′ ˙ By Ptolemy theorem on the cyclic quadrilateral BCQ P , we have that BP · CQ + BC · P Q = BQ CP . √ 2 Substituting the side lengths back gives 4 · 5 + 7 · 7 = AM , so AM = 69 . Solution 2: A ′ ′ Q P P H Q C B M ©2026 HMMT Instead of reflecting across the midpoints of AB, AC , the reflection across the lines AB, AC also works too. Because H is the orthocenter of △ ABC , the circumcircle of △ ABH is the reflection of the circumcircle ′ of △ ABC across lines AB . Let P be the reflection of P across line AB , then it must lie on the ′ circumcircle of △ ABC . Similarly, let Q be the reflection of Q across line AC , then it must lie on the circumcircle of △ ABC too. ′ ′ ′ ′ We note that AP = 4 and AQ = 5 . Moreover, ∠ P AB = ∠ BAM = ∠ M AC = ∠ CAQ , so ′ ′ ′ ′ ∠ P AM = ∠ BAC = ∠ M AQ . As a result, P M = BC = M Q = 7 . ′ ′ Now, we can only focus on quadrilateral AP Q M , which we know all side lengths. There are a couple ways to find the length AM . ′ ′ • Let T be the intersection of lines AM and P Q . Note that ′ ′ ′ 2 (a) Since △ M T P ∼ △ M P A , we have M T · M A = M P = 49 . (This is known as “shooting lemma”.) √ ′ ′ ′ ′ ′ ′ (b) Since △ AP T ∼ △ AM Q , we have AT · AM = AP · AQ = 20 . (This is known as AP · AQ - inversion.) Adding these two equations together implies that 69 = M T · M A + AT · AM = AM ( M T + AT ) = √ 2 AM . Therefore, the length of AM is 69 . ′ ′ • Using the Law of Cosines on △ AP M and △ AQ M , 2 ′ 2 ′ 2 ′ ′ ′ ′ AM = AP + P M − 2 AP · P M cos ∠ AP M = 65 − 56 cos ∠ AP M 2 ′ 2 ′ 2 ′ ′ ′ ′ AM = AQ + Q M − 2 AQ · Q M cos ∠ AQ M = 74 − 70 cos ∠ AQ M ′ ′ ◦ ′ ′ ◦ Moreover, since ∠ AP M + ∠ AQ M = 180 , we get that cos ∠ AP M + cos ∠ AQ M = 0 . Solving √ the two equations above gives us that the length of AM is 69 .