Berkeley Math Circle Monthly Contest 4

Solution:

The answers, expressed in terms of numbers of sides, are $(3,9)$, $(4,6)$, $(5,5)$, and $(8,4)$. They are easily verified to be matching.

To prove that there are no others, first note that if the first polygon has at most five sides, the second one necessarily has one of the numbers of sides necessary to make one of the first three matching pairs. The second polygon cannot be a triangle (since then the first polygon would have an interior angle of $180^{\circ}$), and if it is a square or pentagon, we recover the last two matching pairs. So if there are any matching pairs other than the four listed, both polygons must have at least six sides. But then the interior angle of the first is at least $120^{\circ}$ and the exterior of the second is at most $60^{\circ}$, which is impossible.