HMMT 二月 2006 · 几何 · 第 5 题

HMMT February 2006 — Geometry — Problem 5

Triangle ABC has side lengths AB = 2 5, BC = 1, and CA = 5. Point D is on side AC such that CD = 1, and F is a point such that BF = 2 and CF = 3. Let E be the intersection of lines AB and DF . Find the area of CDEB .

解析

Triangle ABC has side lengths AB = 2 5, BC = 1, and CA = 5. Point D is on side AC such that CD = 1, and F is a point such that BF = 2 and CF = 3. Let E be the intersection of lines AB and DF . Find the area of CDEB . 1 22 Answer: 35 Solution: Draw segment AF . Then notice AF = 4, and we have a right triangle. 4 8 Now draw line CE , let it intersect AF at G . By Ceva, F G = and AG = . Using 3 3 AE [ AEF ] 4 mass points we find that = 6 so = 6 , and since [ ABF ] = 4, [ BEF ] = . EB [ BEF ] 7 1 6 It’s easy to see that [ CDF ] = [ ACF ] = , so 5 5 6 4 22 [ BCDE ] = [ CDF ] − [ BEF ] = − = . 5 7 35