HMMT 二月 2004 · 冲刺赛 · 第 35 题
HMMT February 2004 — Guts Round — Problem 35
题目详情
- [12] There are eleven positive integers n such that there exists a convex polygon with n sides whose angles, in degrees, are unequal integers that are in arithmetic progression. Find the sum of these values of n .
解析
- There are eleven positive integers n such that there exists a convex polygon with n sides whose angles, in degrees, are unequal integers that are in arithmetic progression. Find the sum of these values of n . Solution: 106 The sum of the angles of an n -gon is ( n − 2)180, so the average angle measure is ( n − 2)180 /n . The common difference in this arithmetic progression is at least 1, so the difference between the largest and smallest angles is at least n − 1. So the largest angle is at least ( n − 1) / 2 + ( n − 2)180 /n . Since the polygon is convex, this quantity is no larger than 179: ( n − 1) / 2 − 360 /n ≤ − 1, so that 360 /n − n/ 2 ≥ 1 / 2. Multiplying 2 by 2 n gives 720 − n ≥ n . So n ( n + 1) ≤ 720, which forces n ≤ 26. Of course, since the common difference is an integer, and the angle measures are integers, ( n − 2)180 /n must be an integer or a half integer, so ( n − 2)360 /n = 360 − 720 /n is an integer, and then 720 /n must be an integer. This leaves only n = 3 , 4 , 5 , 6 , 8 , 9 , 10 , 12 , 15 , 16 , 18 , 20 , 24 as possibilities. When n is even, ( n − 2)180 /n is not an angle of the polygon, but the mean of the two middle angles. So the common difference is at least 2 when ( n − 2)180 /n is an integer. For n = 20, the middle angle is 162, so the largest angle is at least 162 + 38 / 2 = 181, since 38 is no larger than the difference between the smallest and largest angles. For n = 24, the middle angle is 165, again leading to a contradiction. So no solution exists for n = 20 , 24. All of the others possess solutions: n angles 3 59 , 60 , 61 4 87 , 89 , 91 , 93 5 106 , 107 , 108 , 109 , 110 6 115 , 117 , 119 , 121 , 123 , 125 8 128 , 130 , 132 , 134 , 136 , 138 , 140 , 142 9 136 , . . . , 144 10 135 , 137 , 139 , . . . , 153 12 139 , 141 , 143 , . . . , 161 15 149 , 150 , . . . , 163 16 150 , 151 , . . . , 165 18 143 , 145 , . . . , 177 (These solutions are quite easy to construct.) The desired value is then 3 + 4 + 5 + 6 + 8 + 9 + 10 + 12 + 15 + 16 + 18 = 106. 11