HMMT 二月 2009 · 几何 · 第 8 题

HMMT February 2009 — Geometry — Problem 8

[ 7 ] Triangle ABC has side lengths AB = 231 , BC = 160 , and AC = 281 . Point D is constructed on the opposite side of line AC as point B such that AD = 178 and CD = 153 . Compute the distance from B to the midpoint of segment AD.

解析

[ 7 ] Triangle ABC has side lengths AB = 231 , BC = 160 , and AC = 281 . Point D is constructed on the opposite side of line AC as point B such that AD = 178 and CD = 153 . Compute the distance from B to the midpoint of segment AD. Answer: 208 Solution: Note that ∠ ABC is right since 2 2 2 2 BC = 160 = 50 · 512 = ( AC − AB ) · ( AC + AB ) = AC − AB . ′ ′ ′ ′ Construct point B such that ABCB is a rectangle, and construct D on segment B C such that ′ AD = AD . Then ′ ′ 2 ′ 2 ′ 2 2 2 2 B D = AD − AB = AD − BC = ( AD − BC )( AD + BC ) = 18 · 338 = 78 . ′ ′ ′ ′ ′ It follows that CD = B C − B D = 153 = CD ; thus, points D and D coincide, and AB ‖ CD. Let M denote the midpoint of segment AD, and denote the orthogonal projections M to lines AB and BC by P and Q respectively. Then Q is the midpoint of BC and AP = 39 , so that P B = AB − AP = 192 and √ √ 2 2 2 2 BM = P Q = 80 + 192 = 16 5 + 12 = 208 .