HMMT 二月 2012 · TEAM1 赛 · 第 4 题

HMMT February 2012 — TEAM1 Round — Problem 4

[ 20 ] Let ABC be a triangle with AB < AC . Let M be the midpoint of BC . Line l is drawn through M so that it is perpendicular to AM , and intersects line AB at point X and line AC at point Y . Prove ◦ that ∠ BAC = 90 if and only if quadrilateral XBY C is cyclic.

解析

[ 20 ] Let ABC be a triangle with AB < AC . Let M be the midpoint of BC . Line l is drawn through M so that it is perpendicular to AM , and intersects line AB at point X and line AC at point Y . Prove ◦ that ∠ BAC = 90 if and only if quadrilateral XBY C is cyclic. Answer: see below Team A p A P l Y B C M X First, note that XBY C cyclic is equivalent to ∡ BXM = ∡ ACB . However, note that ∡ BXM = ◦ ◦ 90 − ∡ BAM , so XBY C cyclic is in turn equivalent to ∡ BAM + ∡ ACB = 90 . Let the line tangent to the circumcircle of △ ABC at A be p , and let P be an arbitrary point on p on ◦ the same side of AM as B . Note that ∡ P AB = ∡ ACB . If ∡ ACB = 90 − ∡ BAM we have l ⊥ AM and thus the circumcenter O of △ ABC lies on AM . Since AB < AC , we must have O = M , and ◦ ◦ ◦ ◦ ∡ BAC = 90 . Conversely, if ∡ BAC = 90 , ∡ P AM = 90 , and it follows that ∡ ACB = 90 − ∡ BAM .