HMMT 二月 2023 · 冲刺赛 · 第 32 题

HMMT February 2023 — Guts Round — Problem 32

[23] Let ABC be a triangle with ∠ BAC > 90 . Let D be the foot of the perpendicular from A to side BC . Let M and N be the midpoints of segments BC and BD , respectively. Suppose that AC = 2, ∠ BAN = ∠ M AC , and AB · BC = AM . Compute the distance from B to line AM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT February 2023, February 18, 2023 — GUTS ROUND Organization Team Team ID#

解析

[23] Let ABC be a triangle with ∠ BAC > 90 . Let D be the foot of the perpendicular from A to side BC . Let M and N be the midpoints of segments BC and BD , respectively. Suppose that AC = 2, ∠ BAN = ∠ M AC , and AB · BC = AM . Compute the distance from B to line AM . Proposed by: Luke Robitaille, Pitchayut Saengrungkongka √ 285 Answer: 38 Solution: Y A 1 2 2 B C M N D 1 2 H X Extend AM to meet the circumcircle of 4 ABC at X . Then, we have 4 ABM ∼ 4 CXM , which CX AB 1 implies that = . Using the condition AB · BC = AM , we get that CX = . CM AM 2 Now, the key observation is that 4 AN B ∼ 4 ACX . Thus, if we let Y be the reflection of X across √ √ ◦ 2 2 point C , we get that ∠ AY C = 90 . Thus, Pythagorean’s theorem gives AY = AC − CY = 15 / 4 √ √ 2 2 2 and AX = AY + XY = 19 / 4. Finally, note that the distance from B and C to line AM are equal. Thus, let H be the foot from C to AM . Then, from 4 XCH ∼ 4 XAY , we get that √ √ CX 15 1 / 2 285 √ CH = AY · = · = . AX 2 38 19 / 2