HMMT 二月 2024 · 冲刺赛 · 第 16 题

HMMT February 2024 — Guts Round — Problem 16

[9] Let ABC be an acute isosceles triangle with orthocenter H . Let M and N be the midpoints of sides AB and AC , respectively. The circumcircle of triangle M HN intersects line BC at two points X and Y . 2 Given XY = AB = AC = 2 , compute BC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT February 2024, February 17, 2024 — GUTS ROUND Organization Team Team ID#

解析

[9] Let ABC be an acute isosceles triangle with orthocenter H . Let M and N be the midpoints of sides AB and AC , respectively. The circumcircle of triangle M HN intersects line BC at two points 2 X and Y . Given XY = AB = AC = 2 , compute BC . Proposed by: Andrew Wen √ Answer: 2( 17 − 1) Solution: A H M N B C Y X D Let D be the foot from A to BC , also the midpoint of BC . Note that DX = DY = M A = M B = M D = N A = N C = N D = 1 . Thus, M N XY is cyclic with circumcenter D and circumradius 1 . H lies on this circle too, hence DH = 1 . If we let DB = DC = x , then since △ HBD ∼ △ BDA , √ √ 17 − 1 2 2 4 2 2 2 BD = HD · AD = ⇒ x = 4 − x = ⇒ x = 4 − x = ⇒ x = . 2 √ 2 2 2 Our answer is BC = (2 x ) = 4 x = 2( 17 − 1)