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Difference between revisions of "Mock AIME 2 2006-2007 Problems/Problem 12"
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*[[Mock AIME 2 2006-2007/Problem 11 | Previous Problem]]
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*[[Mock AIME 2 2006-2007 Problems/Problem 11 | Previous Problem]]
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*[[Mock AIME 2 2006-2007/Problem 13 | Next Problem]]
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*[[Mock AIME 2 2006-2007 Problems/Problem 13 | Next Problem]]
 
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Revision as of 14:33, 3 April 2012

Problem

In quadrilateral $ABCD,$ $m \angle DAC= m\angle DBC$ and $\frac{[ADB]}{[ABC]}=\frac12.$ $O$ is defined to be the intersection of the diagonals of $ABCD$. If $AD=4,$ $BC=6$, $BO=1,$ and the area of $ABCD$ is $\frac{a\sqrt{b}}{c},$ where $a,b,c$ are relatively prime positive integers, find $a+b+c.$


Note*: $[ABC]$ and $[ADB]$ refer to the areas of triangles $ABC$ and $ADB.$

Solution

$m\angle DAC=m\angle DBC \Rightarrow ABCD$ is a cylic quadrilateral.

Let $DO=a, AO=b$

$\triangle AOD$ ~ $\triangle BOC \Rightarrow b=\frac{2}{3}$

Also, from the Power of a Point Theorem, $DO \cdot BO=AO\cdot CO\Rightarrow CO=\frac{3a}{2}$

Notice $\frac{[AOD]}{[BOC]}=(\frac{2}{3})^2\Rightarrow [BOC]=\frac{9}{4}[AOD]$

It is given $\frac{[AOD]+[AOB]}{[AOB]+[BOC]}=\frac{[ADB]}{[ABC]}=\frac{1}{2} \Rightarrow [AOB]=\frac{[AOD]}{4}$

Note that $\sin{\angle AOB}=\sin{(180-\angle AOD)}=\sin{\angle AOD}$

Then $[AOB]=\frac{\frac{2}{3}\cdot 1\cdot\sin{\angle AOB}}{2}=\frac{\sin{\angle AOD}}{3}$ and $[AOD]=\frac{\frac{2}{3}\cdot a\cdot\sin{\angle AOD}}{2}=\frac{a\sin{\angle AOD}}{3}$

$\Rightarrow a=4$


$[COD]=9[AOD]$

Thus we need to find $[ABCD]=\frac{25}{2}[AOD]$

Note that $\triangle AOD$ is isosceles with sides $4, 4, \frac{2}{3}$ so we can draw the altitude from D to split it to two right triangles.

$[AOD]=\frac{\sqrt{143}}{9}$

Thus $[ABCD]=\frac{25\sqrt{143}}{18}\rightarrow\boxed{186}$

See also


Problem Source

AoPS users 4everwise and Altheman collaborated to create this problem.

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