2010 AMC 12A Problem 25

A convex cyclic quadrilateral is determined up to rotation and translation by its cyclic sequence of side lengths, and it exists exactly when the largest side is less than the sum of the others. With perimeter $32,$ this means each side is at most $15.$ First count ordered quadruples $(a,b,c,d)$ of positive integers with $a+b+c+d=32$ and each entry at most $15.$ Without the upper bound there are $\binom{31}{3}=4495;$ removing those with some entry at least $16$ subtracts $4\binom{16}{3}=2240,$ leaving $2255.$ Rotations of the quadrilateral correspond to cyclic permutations of $(a,b,c,d).$ By Burnside’s lemma the number of distinct quadrilaterals is $$\frac{1}{4}\big(2255+f_1+f_2+f_3\big),$$ where $f_i$ counts quadruples fixed by rotating $i$ positions. A one- or three-step rotation fixes only $(8,8,8,8),$ so $f_1=f_3=1.$ A two-step rotation fixes $(a,b,a,b)$ with $a+b=16$ and $1\le a,b\le15,$ giving $f_2=15.$ Hence the count is $$\begin{aligned} \frac{1}{4}(2255+1+15+1) &= \frac{2272}{4} \\ &= 568. \end{aligned}$$ Thus, C is the correct answer.