HMMT 十一月 2016 · THM 赛 · 第 3 题

HMMT November 2016 — THM Round — Problem 3

The three points A , B , C form a triangle. AB = 4, BC = 5 AC = 6. Let the angle bisector of ∠ A intersect side BC at D . Let the foot of the perpendicular from B to the angle bisector of ∠ A be E . Let the line through E parallel to AC meet BC at F . Compute DF .

解析

The three points A , B , C form a triangle. AB = 4, BC = 5 AC = 6. Let the angle bisector of ∠ A intersect side BC at D . Let the foot of the perpendicular from B to the angle bisector of ∠ A be E . Let the line through E parallel to AC meet BC at F . Compute DF . Proposed by: Allen Liu Answer: 1 / 2 AB AC Since AD bisects ∠ A , by the angle bisector theorem = , so BD = 2 and CD = 3. Extend BE BD CD to hit AC at X . Since AE is the perpendicular bisector of BX , AX = 4. Since B, E, X are collinear, applying Menelaus’ Theorem to the triangle ADC , we have AE DB CX · · = 1 ED BC XA AE DF DE DC 1 This implies that = 5, and since EF ‖ AC , = , so DF = = . ED DC DA 6 2