HMMT 二月 2025 · 几何 · 第 5 题

HMMT February 2025 — Geometry — Problem 5

Let △ ABC be an equilateral triangle with side length 6. Let P be a point inside triangle △ ABC such ◦ that ∠ BP C = 120 . The circle with diameter AP meets the circumcircle of △ ABC again at X ̸ = A . Given that AX = 5, compute XP .

解析

Let △ ABC be an equilateral triangle with side length 6. Let P be a point inside triangle △ ABC such ◦ that ∠ BP C = 120 . The circle with diameter AP meets the circumcircle of △ ABC again at X ̸ = A . Given that AX = 5, compute XP . Proposed by: Pitchayut Saengrungkongka √ √ Answer: 23 − 2 3 Solution: A X P B C ′ A ′ ′ ◦ ′ ◦ Let A be the antipode of A . As ∠ AXA = 90 , we have X , P , and A are collinear. As ∠ BP C = 120 ′ ◦ ◦ ′ and ∠ BA C = 180 − ∠ BAC = 120 , it follows that P lies on the circle with center A passing √ √ ′ BC ′ ′ √ through B and C , so A P = = 2 3 and AA = 2 A B = 4 3. By the Pythagorean theorem, 3 q √ √ √ √ ′ ′ ′ 2 2 XA = (4 3) − 5 = 23, so the answer is XA − P A = 23 − 2 3 .