Difference between revisions of "2010 AMC 12B Problems/Problem 22"

Revision as of 11:00, 4 July 2013

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 $\boxed{\textbf{(D)}}$ .

2010 AMC 12B (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 == See also ==
 {{AMC12 box|year=2010|num-b=21|num-a=23|ab=B}}
+{{MAA Notice}}

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Difference between revisions of "2010 AMC 12B Problems/Problem 22"

Revision as of 11:00, 4 July 2013

Problem 22

Solution

See also