Two quadrilaterals are considered the same if one can be obtained from the other by a rotation and a translation. How many different convex cyclic quadrilaterals are there with integer sides and perimeter equal to 32? $\textbf{(A)}\ 560 \qquad$ $ \textbf{(B)}\ 564 \qquad$ $ \textbf{(C)}\ 568 \qquad$ $ \textbf{(D)}\ 1498 \qquad$ $ \textbf{(E)}\ 2255$
Solution
A convex cyclic quadrilateral with integer sides $a,b,c,d$ exists exactly when each side is less than the sum of the other three, i.e. each side is less than $16$; and the sides in order determine it up to congruence. Rotations and translations are allowed but not reflections, so we count ordered $4$-tuples up to cyclic rotation (Burnside, rotation group of order $4$).
The positive solutions of $a+b+c+d=32$ number $\binom{31}3=4495$. Those with some side at least $16$: for a given side, $\binom{16}3=560$, and two sides cannot both be $\ge16$, so $4\cdot560=2240$ must be removed. This leaves $2255$ tuples fixed by the identity.
A rotation by $180^\circ$ fixes the tuples $(a,b,a,b)$ with $a+b=16$, $1\le a\le15$: $15$ tuples. Each rotation by $90^\circ$ or $270^\circ$ fixes only $(8,8,8,8)$.
The number of classes is $\dfrac{2255+15+1+1}4=568$.
Two quadrilaterals are considered the same if one can be obtained from the other by a rotation and a translation. How many different convex cyclic quadrilaterals are there with integer sides and perimeter equal to 24?
