2010 AMC 12B Problems/Problem 22

Problem

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 $\triangle ABD$ 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 $\boxed{\textbf{(D)} \sqrt{\frac{425}{2}}}$ .

2010 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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All AMC 12 Problems and Solutions

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2010 AMC 12B Problems/Problem 22

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