HMMT 二月 2026 · 几何 · 第 7 题
HMMT February 2026 — Geometry — Problem 7
题目详情
- Let ABC be an isosceles triangle with AB = AC . Points P and Q are located inside triangle ABC ◦ ◦ ◦ such that BP = P Q = QC . Suppose that ∠ BAP = 20 , ∠ P AQ = 46 , and ∠ QAC = 26 . Compute the measure of ∠ AP C .
解析
- Let ABC be an isosceles triangle with AB = AC . Points P and Q are located inside triangle ABC ◦ ◦ ◦ such that BP = P Q = QC . Suppose that ∠ BAP = 20 , ∠ P AQ = 46 , and ∠ QAC = 26 . Compute the measure of ∠ AP C . Proposed by: Andrew Brahms, Jackson Dryg, Jason Mao, Pitchayut Saengrungkongka ◦ Answer: 74 Solution 1: ©2026 HMMT A ′ P P Q C B ′ ′ ∼ Let P be the point outside △ ABC such that △ AP B △ AP C . Note that = ′ ′ ◦ ◦ ◦ • ∠ QAP = ∠ QAC + ∠ CAP = ∠ QAC + ∠ BAP = 20 + 26 = 46 = ∠ P AQ . ′ • AP = AP ′ ′ ∼ Therefore, △ AP Q △ AP Q , so P Q = P Q = QC , which means that Q is the circumcenter of = ′ ′ ′ △ P P C . Since P C = P B = QC , we get that △ QP C is an equilateral triangle. By angle chasing, 1 ′ ′ ◦ ′ ◦ ◦ ◦ ◦ ∠ AP C = ∠ AP P + ∠ P P C = 90 − ∠ P AQ + · ∠ P QC = 90 − 46 + 30 = 74 . 2 Solution 2: A P Q C B X Let X be the reflection of B across AP . Then, note that • AX = AB = AC . ◦ ◦ ◦ • ∠ XAC = ∠ P AQ − ∠ P AX = ∠ P AQ − ∠ P AB = 46 − 20 = 26 = ∠ QAC . ∼ Therefore, △ AXQ △ ACQ , so X is the reflection of C across AQ too. Therefore, P X = P B = P Q = and QX = QC = P Q , so △ P QX is an equilateral triangle. ◦ Note that Q is the circumcenter of △ P XC . Therefore, ∠ P CX = 30 . Because XC ⊥ AQ , we ◦ ◦ ◦ ◦ ◦ have ∠ ACX = 90 − ∠ QAC = 90 − 26 = 64 , so ∠ ACP = ∠ ACX − ∠ P CX = 34 . Finally, ◦ ◦ ◦ ◦ ◦ ∠ AP C = 180 − ∠ ACP − ∠ P AC = 180 − 72 − 34 = 74 . ©2026 HMMT