# Mock AIME 2 2006-2007 Problems/Problem 12

## 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

 Mock AIME 2 2006-2007 (Problems, Source) Preceded byProblem 11 Followed byProblem 13 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

## Problem Source

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

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