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 $\sin{\angle AOD}=\sin{(180-\angle AOD)}$ , $[AOD]=\sin{\angle AOD}\cdot\frac{1}{2}\cdot a\cdot\frac{2}{3}=\frac{a}{3}\cdot\sin{\angle AOD}$ , $[AOB]=\sin{(180-\angle AOD)}\cdot\frac{1}{2}\cdot\frac{2}{3}\cdot 1=\frac{[AOD]}{a}$