PUMaC 2008 · 个人决赛（B 组） · 第 2 题

PUMaC 2008 — Individual Finals (Division B) — Problem 2

Let P be a convex polygon, and let n ≥ 3 be a positive integer. On each side of P , erect a regular n -gon that shares that side of P , and is outside P . If none of the interiors of these regular n -gons overlap, we call P n - good . (a) Find the largest value of n such that every convex polygon is n -good. (b) Find the smallest value of n such that no convex polygon is n -good.

解析

Let P be a convex polygon, and let n ≥ 3 be a positive integer. On each side of P , erect a regular n -gon that shares that side of P , and is outside P . If none of the interiors of these regular n -gons overlap, we call P n - good . (a) Find the largest value of n such that every convex polygon is n -good. (b) Find the smallest value of n such that no convex polygon is n -good. 2 ◦ ( ANS: The angle measure of a regular n -gon is (1 − )180 . A polygon is n -good if and only if n we can fit two of these angles outside each angle of the polygon; that is, if and only if 2 ◦ ◦ 2(1 − )180 + x ≤ 360 for each angle x of the polygon. Simplifying the inequality, we have n ◦ 720 ◦ x ≤ . (a) The answer is 4. For n = 4, we have x ≤ 180 , which is certaintly true for each angle n ◦ ◦ of a convex polygon. For n = 5, we have x ≤ 144 , so no polygon with an angle larger than 144 is 5-good. (b) The answer is 13. An equilateral triangle is n -good for all 3 ≤ n ≤ 12. We now prove no polygon is 13-good. Suppose a polygon has k sides. The sum of the angles of the polygon is ◦ ( k − 2)180 2 ◦ ◦ ( k − 2)180 , so there must be some angle which measures at least = (1 − )180 . Since k k 2 2 ◦ ◦ ◦ k ≥ 3, there must be angle measuring at least (1 − )180 ≥ (1 − )180 = 60 . But the condition k 3 ◦ 720 ◦ for 13-goodness is x ≤ < 60 for all angles x , so no polygon is 13-good. CB: GL) 13