HMMT 二月 2018 · 冲刺赛 · 第 20 题

HMMT February 2018 — Guts Round — Problem 20

[ 10 ] Triangle 4 ABC has AB = 21, BC = 55, and CA = 56. There are two points P in the plane of ◦ 4 ABC for which ∠ BAP = ∠ CAP and ∠ BP C = 90 . Find the distance between them. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT February 2018, February 10, 2018 — GUTS ROUND Organization Team Team ID#

解析

[ 10 ] Triangle 4 ABC has AB = 21, BC = 55, and CA = 56. There are two points P in the plane of ◦ 4 ABC for which ∠ BAP = ∠ CAP and ∠ BP C = 90 . Find the distance between them. Proposed by: Michael Tang √ 5 Answer: 409 2 Let P and P be the two possible points P , with AP < AP . Both lie on the ∠ A -bisector and the 1 2 1 2 circle γ with diameter BC . Let D be the point where the ∠ A -bisector intersects BC , let M be the midpoint of BC , and let X be the foot of the perpendicular from M onto the ∠ A -bisector. Since we know the radius of γ , to compute P P it suffices to compute M X . By the angle bisector 1 2 theorem we find BD = 15 and DC = 40, so Stewart’s theorem gives 2 2 2 15 · 40 · 55 + 55 · AD = 21 · 40 + 56 · 15 = ⇒ AD = 24 . 2 2 2 − 21 +15 +24 1 ◦ Then cos ∠ ADB = = , so ∠ ADB = ∠ M DX = 60 . Since DM = BM − BD = 2 · 15 · 24 2 √ 55 25 25 3 − 15 = , we get M X = DM sin ∠ M DX = . Hence 2 2 4 √ ( ) √ √ 2 √ ( ) 2 √ 55 25 3 5 409 √ P P = 2 − = . 1 2 2 4 2