HMMT 二月 2011 · 冲刺赛 · 第 21 题
HMMT February 2011 — Guts Round — Problem 21
题目详情
- [ 12 ] Let ABCD be a quadrilateral inscribed in a circle with center O . Let P denote the intersection √ of AC and BD . Let M and N denote the midpoints of AD and BC . If AP = 1, BP = 3, DP = 3, and AC is perpendicular to BD , find the area of triangle M ON .
解析
- [ 12 ] Let ABCD be a quadrilateral inscribed in a circle with center O . Let P denote the intersection √ of AC and BD . Let M and N denote the midpoints of AD and BC . If AP = 1, BP = 3, DP = 3, and AC is perpendicular to BD , find the area of triangle M ON . 3 Answer: 4 ◦ ◦ ◦ We first prove that ON P M is a parallelogram. Note that AP D and BP C are both 30 − 60 − 90 ′ ′ ◦ triangles. Let M denote the intersection of M P and BC . Since ∠ BP M = ∠ M P D = 30 , we have M P ⊥ BC . Since ON is the perpendicular bisector of BC , we have M P//N O . Similarly, we have M O//N P . Thus ON P M is a parallelogram. It follows that the area of triangle M ON is equal to the 1 1 3 ◦ area of triangle M P N , which is equal to · 1 · 3 · sin ∠ M P N = · 1 · 3 · sin 150 = . 2 2 4