Harvard-MIT Mathematics Tournament, Team Round A

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 $n \geq 4$. Let $\theta_{1}, \theta_{2}, \ldots, \theta_{n}$ be $n$ angles less than $180^{\circ}$ whose sum equals $180(n-2)$ degrees. The statement is clearly true if $n=4$ and $\theta_{1}=\theta_{3}=180^{\circ}-\theta_{2}=180^{\circ}-\theta_{4}$ since we can easily form a parallelogram, so assume otherwise.

I claim that there exist two adjacent angles whose sum is greater than $180^{\circ}$. Assume otherwise. Then, we have $\theta_{i}+\theta_{i+1} \leq 180$ for $i=1,2, \ldots, n$, where $\theta_{n+1}=\theta_{1}$. Summing these inequalities over all $i$ yields $2 \cdot 180(n-2) \leq 180 n$, which is equivalent to $n \leq 4$. 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 $180^{\circ}$. Without loss of generality, let these angles be $\theta_{n-1}$ and $\theta_{n}$, relabeling if necessary. By the inductive hypothesis, we may construct an $(n-1)$-gon with angles $\theta_{1}, \theta_{2}, \ldots, \theta_{n-2}, \theta_{n-1}+\theta_{n}-180^{\circ}$, as these angles are each less than $180^{\circ}$ and their sum equals $180(n-3)$ degrees. Consider the vertex with angle $\theta_{n-1}+\theta_{n}-180^{\circ}$. Note that we can "clip off" a triangle with angles $\theta_{n-1}+\theta_{n}-180^{\circ}, 180^{\circ}-\theta_{n-1}$, and $180^{\circ}-\theta_{n}$ at this vertex, yielding an $n$-gon with the desired angles, completing the inductive step.