AMC12 2010 12B, Problem 22

Let $ABCD$ be a cyclic quadrilateral. The side lengths of $ABCD$ are distinct integers less than $15$ such that $BC\cdot CD=AB\cdot DA$. What is the largest possible value of $BD$?

Solution: Let $AB = a$, $BC = b$, $CD = c$, and $AD = d$.  We see that by the Law of Cosines on $\triangle ABD$ and $\triangle CBD$, we have: $BD^2 = a^2 + d^2 - 2ad\cos{\angle BAD}$. $BD^2 = b^2 + c^2 - 2bc\cos{\angle BCD}$. We are given that $ad = bc$ and $ABCD$ is a cyclic quadrilateral. As a property of cyclic quadrilaterals, opposite angles are supplementary so $\angle BAD = 180 - \angle BCD$, therefore $\cos{\angle BAD} = -\cos{\angle BCD}$. So, $2ad\cos{\angle BAD} = -2bc\cos{\angle BCD}$. Adding, we get $2BD^2 = a^2 + b^2 + c^2 + d^2$. We now look at the equation $ad = bc$.  Suppose that $a = 14$.  Then, we must have either $b$ or $c$ equal $7$.  Suppose that $b = 7$.  We let $d = 6$ and $c = 12$. $2BD^2 = 196 + 49 + 36 + 144 = 425$, so our answer is $\fbox{\sqrt{\dfrac{425}{2}}}$.

Answer: \sqrt{\dfrac{425}{2}}