smc

It should first be noted that given any quadrilateral of fixed side lengths, there is exactly one way to manipulate the angles so that the quadrilateral becomes cyclic. Proof. Given a quadrilateral $ABCD$ where all sides are fixed (in a certain order), we can construct the diagonal $\overline{BD}$. When $BD$ is the minimum allowed by the triangle inequality, one of the angles $\angle DAB$ or $\angle BCD$ will be degenerate and measure $0^{\circ }$, so opposite angles will sum to less than $180^{\circ }$. When $BD$ is the maximum allowed, one of the angles will be degenerate and measure $180^{\circ }$, so opposite angles will sum to more than $180^{\circ }$. Thus, since the sum of opposite angles increases continuously as $BD$ is lengthened from the minimum to the maximum values, there is a unique value of $BD$ somewhere in the middle such that the sum of opposite angles is exactly $180^{\circ }$. Denote $a$, $b$, $c$, and $d$ as the integer side lengths of the quadrilateral. Without loss of generality, let $a\ge b\ge c\ge d$. Since $a+b+c+d=32$, the Triangle Inequality implies that $a\le 15$. We will now split into $5$ cases. Case $1$: $a=b=c=d$ ($4$ side lengths are equal) Clearly there is only $1$ way to select the side lengths $(8,8,8,8)$, and no matter how the sides are rearranged only $1$ unique quadrilateral can be formed. Case $2$: $a=b=c>d$ or $a>b=c=d$ ($3$ side lengths are equal) If $3$ side lengths are equal, then each of those side lengths can only be integers from $6$ to $10$ except for $8$ (because that is counted in the first case). Obviously there is still only $1$ unique quadrilateral that can be formed from one set of side lengths, resulting in a total of $4$ quadrilaterals. Case $3$: $a=b>c=d$ ($2$ pairs of side lengths are equal) $a$ and $b$ can be any integer from $9$ to $15$, and likewise $c$ and $d$ can be any integer from $1$ to $7$. However, a single set of side lengths can form $2$ different cyclic quadrilaterals (a rectangle and a kite), so the total number of quadrilaterals for this case is $7\cdot 2=14$. Case $4$: $a=b>c>d$ or $a>b=c>d$ or $a>b>c=d$ ($2$ side lengths are equal) If the $2$ equal side lengths are each $1$, then the other $2$ sides must each be $15$, which we have already counted in an earlier case. If the equal side lengths are each $2$, there is $1$ possible set of side lengths. Likewise, for side lengths of $3$ there are $2$ sets. Continuing this pattern, we find a total of $1+2+3+4+4+5+7+5+4+4+3+2+1=45$ sets of side lengths. (Be VERY careful when adding up the total for this case!) For each set of side lengths, there are $3$ possible quadrilaterals that can be formed, so the total number of quadrilaterals for this case is $3\cdot 45=135$. Case $5$: $a>b>c>d$ (no side lengths are equal) Using the same counting principles starting from $a=15$ and eventually reaching $a=9$, we find that the total number of possible side lengths is $69$. There are $4!$ ways to arrange the $4$ side lengths, but there is only $1$ unique quadrilateral for $4$ rotations, so the number of quadrilaterals for each set of side lengths is $\frac{4!}{4}=6$. The total number of quadrilaterals is $6\cdot 69=414$. And so, the total number of quadrilaterals that can be made is $414+135+14+4+1=\boxed{568}$.