HMMT 二月 2009 · 代数 · 第 6 题
HMMT February 2009 — Algebra — Problem 6
题目详情
- [ 5 ] Let x and y be positive real numbers and θ an angle such that θ 6 = n for any integer n . Suppose 2 sin θ cos θ = x y and 4 4 cos θ sin θ 97 sin 2 θ
- = . 4 4 3 3 x y x y + y x y x Compute + . y x
解析
- [ 5 ] Let x and y be positive real numbers and θ an angle such that θ 6 = n for any integer n . Suppose 2 sin θ cos θ = x y and 4 4 cos θ sin θ 97 sin 2 θ
- = . 4 4 3 3 x y x y + y x 1 y x Compute + . y x Answer: 4 Solution: From the first relation, there exists a real number k such that x = k sin θ and y = k cos θ . Then we have 4 4 cos θ sin θ 194 sin θ cos θ
- = = 194 . 4 2 4 2 cos θ sin θ sin θ cos θ (cos θ + sin θ ) 4 4 y x 2 2 cos θ sin θ Notice that if t = + then ( t − 2) − 2 = + = 194 and so t = 4. 4 4 y x sin θ cos θ