HMMT 二月 2018 · 冲刺赛 · 第 13 题

HMMT February 2018 — Guts Round — Problem 13

[ 9 ] Suppose 4 ABC has lengths AB = 5, BC = 8, and CA = 7, and let ω be the circumcircle of 4 ABC . Let X be the second intersection of the external angle bisector of ∠ B with ω , and let Y be the foot of the perpendicular from X to BC . Find the length of Y C .

解析

[ 9 ] Suppose 4 ABC has lengths AB = 5, BC = 8, and CA = 7, and let ω be the circumcircle of 4 ABC . Let X be the second intersection of the external angle bisector of ∠ B with ω , and let Y be the foot of the perpendicular from X to BC . Find the length of Y C . Proposed by: Caleb He 13 Answer: 2 − − → ◦ Extend ray AB to a point D , Since BX is an angle bisector, we have ∠ XBC = ∠ XBD = 180 − ∠ XBA = ∠ XCA , so XC = XA by the inscribed angle theorem. Now, construct a point E on BC ∼ ∼ so that CE = AB . Since ∠ BAX ∠ BCX , we have 4 BAX 4 ECX by SAS congruence. Thus, = = XB = XE , so Y bisects segment BE . Since BE = BC − EC = 8 − 5 = 3, we have Y C = EC + Y E = 1 13 5 + · 3 = . 2 2 (Archimedes Broken Chord Thoerem).