HMMT 十一月 2021 · 冲刺赛 · 第 7 题
HMMT November 2021 — Guts Round — Problem 7
题目详情
- [7] Two unit squares S and S have horizontal and vertical sides. Let x be the minimum distance between 1 2 a point in S and a point in S , and let y be the maximum distance between a point in S and a point 1 2 1 in S . Given that x = 5, the difference between the maximum and minimum possible values for y can be 2 √ written as a + b c , where a , b , and c are integers and c is positive and square-free. Find 100 a + 10 b + c .
解析
- [7] Two unit squares S and S have horizontal and vertical sides. Let x be the minimum distance 1 2 between a point in S and a point in S , and let y be the maximum distance between a point in S and 1 2 1 a point in S . Given that x = 5, the difference between the maximum and minimum possible values 2 √ for y can be written as a + b c , where a , b , and c are integers and c is positive and square-free. Find 100 a + 10 b + c . Proposed by: Daniel Zhu Answer: 472 Solution: Consider what must happen in order for the minimum distance to be exactly 5 . Let one square, say S have vertices of (0 , 0) , (0 , 1) , (1 , 0) , and (1 , 1) . Further, assume WLOG that the center 1 1 1 of S is above the line y = and to the right of the line x = , determined by the center of S . There 2 1 2 2 are three cases to consider: • the right side of S and the left side of S are 5 units apart, and the bottom left vertex of S lies 1 2 2 under the line y = 1; • the top side of S and the bottom side of S are 5 units apart, and the bottom left vertex of S 1 2 2 lies to the left of the line x = 1; √ 2 2 • the bottom left coordinate of S is ( a, b ) with a, b ≥ 1, and 5 = ( a − 1) + ( b − 1) . 2 We see that the first two cases are symmetric, so consider the case where the left edge of S lies on 2 the line x = 6 . When this is true, the maximum distance will be achieved between (0 , 0) and the upper right vertex of S . The upper right vertex can achieve the points (7 , c ) where 1 ≤ c ≤ 2 , and so 2 √ √ y ∈ [ 50 , 53] . The other case we have to consider is when the bottom left vertex of S , ( a, b ) , is above y = 1 and 2 to the right of x = 1 , in which case the maximum distance is achieved from (0 , 0) and the upper √ 2 2 right vertex of S . This distance is ( a + 1) + ( b + 1) , which, by the triangle inequality, is at most 2 √ √ √ √ 2 2 2 2 ( a − 1) + ( b − 1) + 2 + 2 = 5 + 2 2. Since equality holds when a = b = 5 / 2 + 1, the largest √ possible maximum here is 5 + 2 2 , and the difference between the largest and smallest possible values √ √ √ of y is 5 + 2 2 − 50 = 5 − 3 2 .