HMMT 二月 2014 · 几何 · 第 3 题

HMMT February 2014 — Geometry — Problem 3

ABC is a triangle such that BC = 10, CA = 12. Let M be the midpoint of side AC . Given that BM is parallel to the external bisector of \ A , find area of triangle ABC . (Lines AB and AC form two angles, one of which is \ BAC . The external bisector of \ A is the line that bisects the other angle.)

解析

ABC is a triangle such that BC = 10, CA = 12. Let M be the midpoint of side AC . Given that BM is parallel to the external bisector of ∠ A , find area of triangle ABC . (Lines AB and AC form two angles, one of which is ∠ BAC . The external bisector of ∠ A is the line that bisects the other angle.) √ Answer: 8 14 Since BM is parallel to the external bisector of ∠ A = ∠ BAM , it is perpendicular 1 to the angle bisector of ∠ BAM . Thus BA = BM = BC = 6. By Heron’s formula, the area of 2 √ √ ∆ ABC is therefore (14)(8)(4)(2) = 8 14. ◦ ◦ ◦