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

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== Solution ==
 
== Solution ==
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For the second problem, let <math>AB = a</math>, <math>BC = b</math>, <math>CD = c</math>, and <math>AD = d</math>.  We see that by the Law of Cosines on <math>\triangle ABD</math>, we have <math>BD^2 = a^2 + d^2 - 2ad\cos{\angle BAD}</math>.  Also, <math>BD^2 = b^2 + c^2 - 2bc\cos{\angle BCD}</math>.  Now, we know that <math>ad = bc</math>.  Also, because <math>ABCD</math> is a cyclic quadrilateral, we must have that <math>\angle BAD = 180 - \angle BCD</math>, so <math>\cos{\angle BAD} = -\cos{\angle BCD}</math>.  Therefore, <math>2ad\cos{\angle BAD} = -2bc\cos{\angle BCD}</math>.  Now, adding, we have <math>2BD^2 = a^2 + b^2 + c^2 + d^2</math>.
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We now look at the equation <math>ad = bc</math>.  Suppose that <math>a = 14</math>.  Then, we must have either <math>b</math> or <math>c</math> equal <math>7</math>.  Suppose that <math>b = 7</math>.  We let <math>d = 6</math> and <math>c = 12</math>. 
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Now, <math>2BD^2 = 196 + 49 + 36 + 144 = 425</math>, so it is <math>\sqrt{\frac{425}{2}}</math> or <math>\boxed{\textbf{(D)}}</math>.
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2010|num-b=21|num-a=23|ab=B}}
 
{{AMC12 box|year=2010|num-b=21|num-a=23|ab=B}}

Revision as of 19:12, 12 February 2011

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

For the second problem, 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 (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions
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