HMMT 十一月 2020 · 冲刺赛 · 第 18 题
HMMT November 2020 — Guts Round — Problem 18
题目详情
- [10] Suppose Harvard Yard is a 17 × 17 square. There are 14 dorms located on the perimeter of the Yard. If s is the minimum distance between two dorms, the maximum possible value of s can be expressed as √ a − b where a, b are positive integers. Compute 100 a + b . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMO 2020, November 14–21, 2020 — GUTS ROUND Organization Team Team ID#
解析
- [10] Suppose Harvard Yard is a 17 × 17 square. There are 14 dorms located on the perimeter of the Yard. If s is the minimum distance between two dorms, the maximum possible value of s can be √ expressed as a − b where a, b are positive integers. Compute 100 a + b . Proposed by: Hahn Lheem Answer: 602 Solution: If two neighboring dorms are separated by a distance of more than s , we can move them slightly closer together and adjust the other dorms, increasing s . Therefore, in an optimal arrangement, the dorms form an equilateral 14-gon with side length s . By scaling, the problem is now equivalent to finding the smallest a such that there exist 14 vertices on the boundary of an a × a square that form an equilateral 14-gon with side length 1. Such a 14-gon must be centrally symmetric, yielding the following picture: y 1 y 2 x x 1 2 1 1 1 2 2 2 2 We know that x + y = x + y = 1 and x + x + 3 = y + y + 2 = a . Moreover, if these equations 1 2 1 2 1 1 2 2 are satisfied, then such a 14-gon exists. We now consider the vectors ~ v = ( x , y ) and ~ v = ( x , y ). 1 1 1 2 2 2 These unit vectors are in the first quadrant and add to ( a − 3 , a − 2), which lies on the line y = x + 1. Since ~ v and ~ v must lie on the first quadrant, from the above diagram we deduce that the minimum 1 2 √ √ 2 2 value of a occurs when one of ~ v , ~ v is (0 , 1), meaning that ( a − 3 , a − 2) = ( , + 1). This means 1 2 2 2 √ 2 that a = + 3, so the maximum possible value of s is 2 √ 2 √ 3 − 17 2 √ = 17 · = 6 − 2 . 2 17 / 2
- 3 2