HMMT 二月 2009 · 几何 · 第 8 题

英文解析

8. [ 7 ] Triangle ABC has side lengths AB = 231 , BC = 160 , and AC = 281 . Point D is constructed onthe opposite side of line AC as point B such that AD = 178 and CD = 153 . Compute the distancefrom 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 BCby 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 .