HMMT 二月 2012 · 几何 · 第 1 题
HMMT February 2012 — Geometry — Problem 1
题目详情
- ABC is an isosceles triangle with AB = 2 and ∡ ABC = 90 . D is the midpoint of BC and E is on AC such that the area of AEDB is twice the area of ECD . Find the length of DE . ◦
解析
- ABC is an isosceles triangle with AB = 2 and ∡ ABC = 90 . D is the midpoint of BC and E is on AC such that the area of AEDB is twice the area of ECD . Find the length of DE . √ 17 Let F be the foot of the perpendicular from E to BC . We have [ AEDB ] + [ EDC ] = Answer: 3 2 1 4 [ ABC ] = 2 ⇒ [ EDC ] = . Since we also have [ EDC ] = ( EF )( DC ) , we get EF = F C = . So 3 2 3 √ 1 17 F D = , and ED = by the Pythagorean Theorem. 3 3 ◦