HMMT 二月 2006 · TEAM1 赛 · 第 9 题
HMMT February 2006 — TEAM1 Round — Problem 9
题目详情
- [40] Let n ≥ 3 be a positive integer. Prove that given any n angles 0 < θ , θ , . . . , θ < 180 , such that 1 2 n their sum is 180( n − 2) degrees, there exists a convex n -gon having exactly those angles, in that order.
解析
- [40] Let n ≥ 3 be a positive integer. Prove that given any n angles 0 < θ , θ , . . . , θ < 1 2 n ◦ 180 , such that their sum is 180( n − 2) degrees, there exists a convex n -gon having exactly those angles, in that order. Solution: We induct on n . The statement holds trivially for n = 3, as all triangles are convex. Now, suppose that the statement is true for n − 1, where n ≥ 4. Let ◦ θ , θ , . . . , θ be n angles less than 180 whose sum equals 180( n − 2) degrees. The 1 2 n ◦ ◦ statement is clearly true if n = 4 and θ = θ = 180 − θ = 180 − θ since we can 1 3 2 4 easily form a parallelogram, so assume otherwise. ◦ I claim that there exist two adjacent angles whose sum is greater than 180 . Assume otherwise. Then, we have θ + θ ≤ 180 for i = 1 , 2 , . . . , n , where θ = θ . Summing i i +1 n +1 1 these inequalities over all i yields 2 · 180( n − 2) ≤ 180 n , which is equivalent to n ≤ 4. Of course, we can have n = 4 if and only if we have equality in each of the above inequalities, forcing us to have a parallelogram contrary to our assumption. ◦ Hence, we have two adjacent angles with sum greater than 180 . Without loss of generality, let these angles be θ and θ , relabeling if necessary. By the inductive n − 1 n ◦ hypothesis, we may construct an ( n − 1)-gon with angles θ , θ , . . . , θ , θ + θ − 180 , 1 2 n − 2 n − 1 n ◦ as these angles are each less than 180 and their sum equals 180( n − 3) degrees. Consider ◦ the vertex with angle θ + θ − 180 . Note that we can “clip off” a triangle with n − 1 n ◦ ◦ ◦ angles θ + θ − 180 , 180 − θ , and 180 − θ at this vertex, yielding an n -gon n − 1 n n − 1 n with the desired angles, completing the inductive step.