HMMT 十一月 2012 · 冲刺赛 · 第 28 题

HMMT November 2012 — Guts Round — Problem 28

[ 15 ] Find the set of all possible values that can be attained by the expression , where a and 2 2 a + b b are positive real numbers. Express your answer in interval notation. 4

解析

[ 15 ] √ 2 1+ 2 ab + b Answer: (0 , ] Suppose that k = for some positive real a, b . We claim that k lies in 2 2 2 a + b √ a 2 +1 1+ 2 a ab + b x +1 2 b (0 , ]. Let x = . We have that = = . Thus, x + 1 = k ( x + 1), so the 2 2 2 2 a 2 b a + b x +1 +1 ( ) b 2 quadratic kt − t + k − 1 = 0 has a positive real root. Thus, its discriminant must be nonnegative, so √ √ 1 − 2 1+ 2 2 2 1 ≥ 4( k − 1)( k ) = ⇒ (2 k − 1) ≤ 2, which implies ≤ k ≤ . Since x > 0, we also have 2 2 √ 1+ 2 k > 0, so we know that k must lie in (0 , ]. 2 √ 1+ 2 2 Now, take any k in the interval (0 , ]. We thus know that 1 ≥ 4 k ( k − 1), so the quadratic 2 √ 1+ 1 − 4 k ( k − 1) 2 2 kt − t + k − 1 = 0 has a positive solution, . Call this solution x . Then k ( x + 1) = x + 1, 2 k 2 x +1 ab + b so = k . If we set a = x and b = 1, we get that = k . Thus, the set of all attainable values 2 2 2 x +1 a + b √ 2 ab + b 1+ 2 of is the interval (0 , ]. 2 2 a + b 2