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Mock AIME 2 2006-2007/Problem 12

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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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