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 $BD$? $\textbf{(A)}\ \sqrt{\dfrac{325}{2}} \qquad$ $ \textbf{(B)}\ \sqrt{185} \qquad$ $ \textbf{(C)}\ \sqrt{\dfrac{389}{2}} \qquad$ $ \textbf{(D)}\ \sqrt{\dfrac{425}{2}} \qquad$ $ \textbf{(E)}\ \sqrt{\dfrac{533}{2}}$
Solution
Let $AB=a$, $BC=b$, $CD=c$, $DA=d$, so $bc=ad$. Since $ABCD$ is cyclic, $\cos C=-\cos A$. The law of cosines in $\triangle ABD$ and $\triangle CBD$ gives $BD^2=a^2+d^2-2ad\cos A$ and $BD^2=b^2+c^2+2bc\cos A$.
Adding the two equations and using $ad=bc$ (so the cosine terms cancel) gives $2BD^2=a^2+b^2+c^2+d^2$, i.e. $BD^2=\dfrac{a^2+b^2+c^2+d^2}2$.
We need distinct integers less than $15$ with $ad=bc$ and the largest sum of squares. The numbers $13$ and $11$ are prime, so they cannot occur. If $14$ occurs, then $7$ must divide one of the other factors, so the numbers are $\{14,7,x,2x\}$ (since $14x=7\cdot2x$), $x=2,\dots,6$; the sum of squares is largest for $x=6$: $196+49+36+144=425$. Without $14$ all numbers are at most $12$ and $11$ is excluded, so the sum of squares is at most $12^2+10^2+9^2+8^2=389<425$.
The best set is $AB=6$, $BC=7$, $CD=12$, $DA=14$ ($6\cdot14=7\cdot12$); a cyclic quadrilateral with these sides exists since $14<6+7+12$. So $BD^2=\dfrac{425}2$, $BD=\sqrt{\dfrac{425}2}$.
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 $BD$?
