2010 AMC 12B Problems/Problem 22

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 $BD$?

$\textbf{(A)}\ \sqrt{\dfrac{325}{2}} \qquad \textbf{(B)}\ \sqrt{185} \qquad \textbf{(C)}\ \sqrt{\dfrac{389}{2}} \qquad \textbf{(D)}\ \sqrt{\dfrac{425}{2}} \qquad \textbf{(E)}\ \sqrt{\dfrac{533}{2}}$

Solution

Let $AB = a$, $BC = b$, $CD = c$, and $AD = d$. We see that by the Law of Cosines on $\triangle ABD$, we have $BD^2 = a^2 + d^2 - 2ad\cos{\angle BAD}$. Also, $BD^2 = b^2 + c^2 - 2bc\cos{\angle BCD}$. Now, we know that $ad = bc$. Also, because $ABCD$ is a cyclic quadrilateral, we must have that $\angle BAD = 180 - \angle BCD$, so $\cos{\angle BAD} = -\cos{\angle BCD}$. Therefore, $2ad\cos{\angle BAD} = -2bc\cos{\angle BCD}$. Now, adding, we have $2BD^2 = a^2 + b^2 + c^2 + d^2$.

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 $c = 12$.

Now, $2BD^2 = 196 + 49 + 36 + 144 = 425$, so it is $\sqrt{\frac{425}{2}}$ or $\boxed{\textbf{(D)}}$.

See also

2010 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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All AMC 12 Problems and Solutions

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