HMMT 十一月 2020 · 团队赛 · 第 10 题
HMMT November 2020 — Team Round — Problem 10
题目详情
- [60] Let x and y be non-negative real numbers that sum to 1. Compute the number of ordered pairs a b a b 1 − a − b ( a, b ) with a, b ∈ { 0 , 1 , 2 , 3 , 4 } such that the expression x y + y x has maximum value 2 .
解析
- [60] Let x and y be non-negative real numbers that sum to 1. Compute the number of ordered pairs a b a b 1 − a − b ( a, b ) with a, b ∈ { 0 , 1 , 2 , 3 , 4 } such that the expression x y + y x has maximum value 2 . Proposed by: Shengtong Zhang Answer: 17 a b a b 1 − a − b 1 1 Solution: Let f ( x, y ) = x y + y x . Observe that 2 is merely the value of f ( , ), so this value 2 2 is always achievable. We claim (call this result ( > )) that if ( a, b ) satisfies the condition, so does ( a + 1 , b + 1). To see 1 1 − a − b a +1 b +1 this, observe that if f ( x, y ) ≤ 2 , then multiplying by the inequality xy ≤ yields x y + 4 a +1 b +1 − 1 − a − b y x ≤ 2 , as desired. For the rest of the solution, without loss of generality we consider the a ≥ b case. If a = b = 0, then f ( x, y ) = 2, so (0 , 0) works. If a = 1 and b = 0, then f ( x, y ) = x + y = 1, so (1 , 0) works. For a ≥ 2, 1 − a ( a, 0) fails since f (1 , 0) = 1 > 2 . 2 2 1 1 If a = 3 and b = 1, f ( x, y ) = xy ( x + y ) = xy (1 − 2 xy ), which is maximized at xy = ⇐⇒ x = y = , 4 2 3 3 3 so (3 , 1) works. However, if a = 4 and b = 1, f ( x, y ) = xy ( x + y ) = xy (( x + y ) − 3 xy ( x + y )) = 1 xy (1 − 3 xy ), which is maximized at xy = . Thus (4 , 1) does not work. 6 From these results and ( > ), we are able to deduce all the pairs that do work ( ↙ represents those pairs that work by ( > )): 4 ↙ ↙ ↙ 7 7 3 ↙ ↙ ↙ 7 3 2 ↙ ↙ ↙ ↙ 7 1 ↙ ↙ 7 3 3 0 7 7 7 3 3 0 1 2 3 4