HMMT 十一月 2025 · 冲刺赛 · 第 30 题

HMMT November 2025 — Guts Round — Problem 30

[15] Point P lies inside triangle ABC such that BP = P C and ∠ AP C − ∠ AP B = 60 . Given that AP = 12, AB = 20, and AC = 25, compute BC . © 2025 HMMT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT November 2025, November 08, 2025 — GUTS ROUND Organization Team Team ID#

解析

[15] Point P lies inside triangle ABC such that BP = P C and ∠ AP C − ∠ AP B = 60 . Given that AP = 12, AB = 20, and AC = 25, compute BC . Proposed by: Jason Mao 75 Answer: = 18 . 75 4 ′ Solution: Let A be the reflection of A about the perpendicular bisector of BC . ′ A A P B C ′ ′ Lemma 1. Quadrilateral ABCA is an isosceles trapezoid, and △ AP A is equilateral. ′ ′ Proof. Observe A C is the reflection of AB about the perpendicular bisector of BC , so ABCA must be an isosceles trapezoid. In particular, P lies on this perpendicular bisector, so ′ ′ ′ ◦ P A = P A and ∠ AP A = ∠ AP C − ∠ A P C = ∠ AP C − ∠ AP B = 60 . ′ Thus △ AP A is equilateral. ′ ′ ′ Since △ AP A is equilateral, we have AA = 12. Then Ptolemy’s Theorem on ABCA yields ′ ′ ′ 2 2 AA · BC + AB · A C = AC · A B = ⇒ 12 · BC + 20 = 25 . 2 2 25 − 20 75 So BC = = . 12 4