HMMT 二月 2012 · 几何 · 第 4 题
HMMT February 2012 — Geometry — Problem 4
题目详情
- There are circles ω and ω . They intersect in two points, one of which is the point A . B lies on ω 1 2 1 such that AB is tangent to ω . The tangent to ω at B intersects ω at C and D , where D is the closer 2 1 2 to B . AD intersects ω again at E . If BD = 3 and CD = 13, find EB/ED . 1
解析
- There are circles ω and ω . They intersect in two points, one of which is the point A . B lies on ω 1 2 1 such that AB is tangent to ω . The tangent to ω at B intersects ω at C and D , where D is the closer 2 1 2 to B . AD intersects ω again at E . If BD = 3 and CD = 13, find EB/ED . 1 √ Answer: 4 3 / 3 [diagram] √ √ √ By power of a point, BA = BD · BC = 4 3. Also, DEB ∼ DBA , so EB/ED = BA/BD = 4 3 / 3.