jmc

As with solution $1$ we would like to note that given any quadrilateral we can change its angles to make a cyclic one. Let $a\ge b\ge c\ge d$ be the sides of the quadrilateral. There are $\left(\genfrac{}{}{0ex}{}{31}{3}\right)$ ways to partition $32$. However, some of these will not be quadrilaterals since they would have one side bigger than the sum of the other three. This occurs when $a\ge 16$. For $a=16$, $b+c+d=16$. There are $\left(\genfrac{}{}{0ex}{}{15}{2}\right)$ ways to partition $16$. Since $a$ could be any of the four sides, we have counted $4\left(\genfrac{}{}{0ex}{}{15}{2}\right)$ degenerate quadrilaterals. Similarly, there are $4\left(\genfrac{}{}{0ex}{}{14}{2}\right)$, $4\left(\genfrac{}{}{0ex}{}{13}{2}\right)\cdots 4\left(\genfrac{}{}{0ex}{}{2}{2}\right)$ for other values of $a$. Thus, there are $\left(\genfrac{}{}{0ex}{}{31}{3}\right)-4\left(\left(\genfrac{}{}{0ex}{}{15}{2}\right)+\left(\genfrac{}{}{0ex}{}{14}{2}\right)+\cdots +\left(\genfrac{}{}{0ex}{}{2}{2}\right)\right)=\left(\genfrac{}{}{0ex}{}{31}{3}\right)-4\left(\genfrac{}{}{0ex}{}{16}{3}\right)=2255$ non-degenerate partitions of $32$ by the hockey stick theorem. We then account for symmetry. If all sides are congruent (meaning the quadrilateral is a square), the quadrilateral will be counted once. If the quadrilateral is a rectangle (and not a square), it will be counted twice. In all other cases, it will be counted 4 times. Since there is $1$ square case, and $7$ rectangle cases, there are $2255-1-2\cdot 7=2240$ quadrilaterals counted 4 times. Thus there are $1+7+\frac{2240}{4}=\boxed{568}$ total quadrilaterals.