Let $ABCD$ be a cyclic quadrilateral. The side lengths of $ABCD$ are distinct integers less than $13$ such that $BC\cdot CD=AB\cdot DA$. What is the largest possible value of $BD$?

1. As in the main problem, $BD^2=\dfrac{AB^2+BC^2+CD^2+DA^2}2$ whenever $BC\cdot CD=AB\cdot DA$.
2. Now the sides are distinct integers at most $12$. The number $11$ is prime and cannot occur. The possible sets $\{a,d;b,c\}$ with $ad=bc$ and the largest sums of squares are $6\cdot12=8\cdot9$ (sum $325$), $5\cdot12=6\cdot10$ (sum $305$), $4\cdot12=6\cdot8$ (sum $260$) and $3\cdot12=4\cdot9$ (sum $250$); the remaining cases give less.
3. The maximum is $325$, attained by $(AB,BC,CD,DA)=(6,8,9,12)$, so $BD=\sqrt{\dfrac{325}2}$.