HMMT 十一月 2020 · THM 赛 · 第 7 题
HMMT November 2020 — THM Round — Problem 7
题目详情
- While waiting for their food at a restaurant in Harvard Square, Ana and Banana draw 3 squares , , on one 1 2 3 of their napkins. Starting with Ana, they take turns filling in the squares with integers from the set { 1 , 2 , 3 , 4 , 5 } such that no integer is used more than once. Ana’s goal is to minimize the minimum value M that the polynomial 2 a x + a x + a attains over all real x , where a , a , a are the integers written in , , respectively. Banana 1 2 3 1 2 3 1 2 3 aims to maximize M . Assuming both play optimally, compute the final value of 100 a + 10 a + a . 1 2 3
解析
- While waiting for their food at a restaurant in Harvard Square, Ana and Banana draw 3 squares , , on one of their napkins. Starting with Ana, they take turns filling in the squares with 1 2 3 integers from the set { 1 , 2 , 3 , 4 , 5 } such that no integer is used more than once. Ana’s goal is to 2 minimize the minimum value M that the polynomial a x + a x + a attains over all real x , where 1 2 3 a , a , a are the integers written in , , respectively. Banana aims to maximize M . Assuming 1 2 3 1 2 3 both play optimally, compute the final value of 100 a + 10 a + a . 1 2 3 Proposed by: Sheldon Kieren Tan Answer: 451 2 − b b Solution: Relabel a , a , a as a, b, c . This is minimized at x = , so M = c − . 1 2 3 2 a 4 a 2 b If in the end a = 5 or b ∈ { 1 , 2 } , then ≤ 1 and M ≥ 0. The only way for Ana to block this is to set 4 a b = 5, which will be optimal if we show that it allows Ana to force M < 0, which we will now do. At this point, Banana has two choices: • If Banana fixes a value of a , Ana’s best move is to pick c = 1, or c = 2 if it has not already been used. The latter case yields M < − 1, while the optimal move in the latter case ( a = 4) yields 25 M = 1 − > − 1. 16 25 • If Banana fixes a value of c , then if that a value is not 1 Ana can put a = 1, yielding M ≤ 4 − < 4 − 1. On the other hand, if Banana fixes c = 1 then Ana’s best move is to put a = 2, yielding 25 M = 1 − < − 1. 8 25 Thus Banana’s best move is to set a = 4, eliciting a response of c = 1. Since 1 − < 0, this validates 16 our earlier claim that b = 5 was the best first move.