HMMT 十一月 2025 · 冲刺赛 · 第 13 题

HMMT November 2025 — Guts Round — Problem 13

[9] Let P be a point and ℓ be a line in the coordinate plane. • If point P were reflected across ℓ and then translated by (+0 , +6), the result would be point A . • If point P were translated by (+0 , +6) and then reflected across ℓ , the result would be point B . Given that AB = 10, compute the maximum possible area of triangle P AB .

解析

[9] Let P be a point and ℓ be a line in the coordinate plane. • If point P were reflected across ℓ and then translated by (+0 , +6), the result would be point A . • If point P were translated by (+0 , +6) and then reflected across ℓ , the result would be point B . Given that AB = 10, compute the maximum possible area of triangle P AB . Proposed by: Jason Mao √ Answer: 5 11 Solution: Let P be the translation of P by (+0 , +6), and let P be the reflection of P about ℓ . 1 2 B A P 1 P 2 P ℓ Lemma 1. We have [ P AB ] = [ P AB ]. 2 © 2025 HMMT Proof. First note that P P AP is an isosceles trapezoid, as segment AP is the reflection of segment 1 2 2 P P over ℓ . Also, P P BP is a parallelogram, as P and B are 6 units above P and P , respectively. 1 1 2 1 2 Thus, both P A and P B are parallel to P P , so P P ∥ AB . 1 1 2 2 Thus, the heights from P and P onto side AB have the same length, so △ P AB and △ P AB have 2 2 heights and bases of the same length, meaning their areas must be the same too. We now compute the area of △ P AB as follows: 2 • Since P and A are reflections of P and P about ℓ , respectively, we have P A = P P = 6. 2 1 2 1 • Also, we are given P B = 6 and AB = 10. 2 √ √ 2 2 Thus, △ P AB is isosceles with base length 10 and height 6 − 5 = 11, so 2 √ √ 1 [ P AB ] = [ P AB ] = · 10 · 11 = 5 11 . 2 2