HMMT 二月 2026 · 冲刺赛 · 第 28 题

HMMT February 2026 — Guts Round — Problem 28

[14] Let ABC be a triangle such that ∠ BAC = 105 , AB = 12 2 , and AC = 17 . Let P be a point such ◦ that P and A lie on different sides of line BC , and ∠ AP B = ∠ AP C = 60 . Compute AP . ©2026 HMMT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT February 2026, February 14, 2026 — GUTS ROUND Organization Team Team ID#

解析

[14] Let ABC be a triangle such that ∠ BAC = 105 , AB = 12 2 , and AC = 17 . Let P be a point ◦ such that P and A lie on different sides of line BC , and ∠ AP B = ∠ AP C = 60 . Compute AP . Proposed by: Sarunyu Thongjarast √ 136 3 Answer: 13 Solution: ©2026 HMMT A O 1 O 2 B C P Let O and O be the circumcenters of triangles ABP and ACP , respectively. Then AO B and AO C 1 2 1 2 ◦ ◦ are isosceles triangles with ∠ AO B = ∠ AO C = 2 · 60 = 120 . Thus, 1 2 √ 1 12 2 AO = √ · AB = √ 1 3 3 and 1 17 AO = √ · AC = √ . 2 3 3 Furthermore, ◦ ◦ ◦ ◦ ∠ O AO = ∠ BAC − ∠ BAO − ∠ O AC = 105 − 30 − 30 = 45 , 1 2 1 2 so ( ) √ ( ) ( ) 1 1 12 2 17 1 [ AO O ] = ( AO )( AO ) sin ∠ O AO = √ √ √ = 34 . 1 2 1 2 1 2 2 2 3 3 2 By law of cosines on triangle AO O , 1 2 √ ( ) ( ) √ √ 2 √ ( ) ( ) ( ) 2 √ 12 2 17 1 12 2 17 13 √ √ √ √ √ √ √ O O = + − 2 = . 1 2 3 3 2 3 3 3 It follows that the distances from A to O O is 1 2 √ [ AO O ] 34 68 3 1 2 2 · = 2 · √ = . O O 13 13 / 3 1 2 Since O A = O P and O A = O P , we know P is the reflection of A over line O O . Thus, AP is 1 1 2 2 1 2 √ 136 3 twice the distance from A to O O , or . 1 2 13