HMMT 二月 2006 · 几何 · 第 5 题
HMMT February 2006 — Geometry — Problem 5
题目详情
英文原题
- 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 theintersection of lines AB and DF . Find the area of CDEB .
解析
英文解析
- 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 theintersection of lines AB and DF . Find the area of CDEB .
Answer:221
Solution: Draw segment AF . Then notice AF = 4, and we have a right triangle.35
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