, 2018

Solution:

The side lengths of the polygon can only be from the set $\{1, \sqrt{2}, \sqrt{3}, 2\}$, which take up $60^{\circ}, 90^{\circ}, 120^{\circ}, 180^{\circ}$ of the circle respectively. By working modulo $60$ degrees we see that $\sqrt{2}$ must be used an even number of times. We now proceed to casework on the longest side of the polygon.

Case 1: If the longest side has length $2$, then the remaining sides must contribute the remaining $180$ degrees. There are $3$ possibilities: $(1,1,1,2), (1, \sqrt{3}, 2), (\sqrt{2}, \sqrt{2}, 2)$.

Case 2: If the longest side has length $\sqrt{3}$, then it takes up either $120^{\circ}$ or $240^{\circ}$ of the circle. In the former case we have $6$ possibilities: $(1,1,1,1, \sqrt{3}), (1, \sqrt{2}, \sqrt{2}, \sqrt{3}), (\sqrt{2}, 1, \sqrt{2}, \sqrt{3}), (1,1, \sqrt{3}, \sqrt{3}), (1, \sqrt{3}, 1, \sqrt{3}), (\sqrt{3}, \sqrt{3}, \sqrt{3})$. In the latter case there is only $1$ possibility: $(1,1, \sqrt{3})$.

Case 3: If the longest side has length $\sqrt{2}$, then it shows up either twice or four times. In the former case we have $2$ possibilities: $(1,1,1, \sqrt{2}, \sqrt{2}), (1,1, \sqrt{2}, 1, \sqrt{2})$. In the latter case there is only $1$ possibility: $(\sqrt{2}, \sqrt{2}, \sqrt{2}, \sqrt{2})$.

Case 4: If all sides have length $1$, then there is $1$ possibility: $(1,1,1,1,1,1)$.

Adding up all cases, we have $3+6+1+2+1+1=14$ polygons.