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

(Created page with '== Problem 22 == Let <math>ABCD</math> be a cyclic quadralateral. The side lengths of <math>ABCD</math> are distinct integers less than <math>15</math> such that <math>BC\cdot CD…')
 
 
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== Problem 22 ==
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== Problem ==
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Let <math>ABCD</math> be a cyclic quadralateral. The side lengths of <math>ABCD</math> are distinct integers less than <math>15</math> such that <math>BC\cdot CD=AB\cdot DA</math>. What is the largest possible value of <math>BD</math>?
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Let <math>ABCD</math> be a cyclic quadrilateral. The side lengths of <math>ABCD</math> are distinct integers less than <math>15</math> such that <math>BC\cdot CD=AB\cdot DA</math>. What is the largest possible value of <math>BD</math>?
  
 
<math>\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}}</math>
 
<math>\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}}</math>
  
 
== Solution ==
 
== Solution ==
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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> and <math>\triangle CBD</math>, we have:
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<math>BD^2 = a^2 + d^2 - 2ad\cos{\angle BAD}</math>. 
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<math>BD^2 = b^2 + c^2 - 2bc\cos{\angle BCD}</math>.
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We are given that <math>ad = bc</math> and <math>ABCD</math> is a cyclic quadrilateral. As a property of cyclic quadrilaterals, opposite angles are supplementary so <math>\angle BAD = 180 - \angle BCD</math>, therefore <math>\cos{\angle BAD} = -\cos{\angle BCD}</math>. So, <math>2ad\cos{\angle BAD} = -2bc\cos{\angle BCD}</math>.
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Adding, we get <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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<math>2BD^2 = 196 + 49 + 36 + 144 = 425</math>, so our answer is <math>\boxed{\textbf{(D)} \sqrt{\frac{425}{2}}}</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}}
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{{MAA Notice}}

Latest revision as of 15:43, 15 February 2021

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

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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