HMMT 二月 2006 · TEAM1 赛 · 第 9 题

英文解析

9. [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 havingexactly those angles, in that order.
   Solution: We induct on n . The statement holds trivially for n = 3, as all trianglesare 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 . Assumeotherwise. Then, we have θ + θ ≤ 180 for i = 1 , 2 , . . . , n , where θ = θ . Summingi 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 aboveinequalities, forcing us to have a parallelogram contrary to our assumption.
   °
   Hence, we have two adjacent angles with sum greater than 180 . Without loss ofgenerality, let these angles be θ and θ , relabeling if necessary. By the inductiven − 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 withn − 1 n
   ° ° °
   angles θ + θ − 180 , 180 − θ , and 180 − θ at this vertex, yielding an n -gonn − 1 n n − 1 nwith the desired angles, completing the inductive step.