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Difference between revisions of "Mock AIME 2 2006-2007 Problems/Problem 12"

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==Solution==
 
==Solution==
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{{solution}}
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<math>m\angle DAC=m\angle DBC \Rightarrow ABCD</math> is a cylic quadrilateral.
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Let <math>DO=a, AO=b</math>
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<math>\triangle AOD</math> ~ <math>\triangle BOC \Rightarrow b=\frac{2}{3}</math>
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Also, from the Power of a Point Theorem, <math>DO \cdot BO=AO\cdot CO\Rightarrow CO=\frac{3a}{2}</math>
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Notice <math>\sin{\angle AOD}=\sin{(180-\angle AOD)}</math> , <math>[AOD]=\sin{\angle AOD}\cdot\frac{1}{2}\cdot a\cdot\frac{2}{3}=\frac{a}{3}\cdot\sin{\angle AOD}</math> , <math>[AOB]=\sin{(180-\angle AOD)}\cdot\frac{1}{2}\cdot\frac{2}{3}\cdot 1=\frac{[AOD]}{a}</math>
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It is given <math>\frac{[AOD]+[AOB]}{[AOB]+[BOC]}=\frac{[ADB]}{[ABC]}=\frac{1}{2} \Rightarrow</math>
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<math>[AOB]=\frac{[AOD]}{4}</math>
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<math>[BOC]=\frac{9}{4}[AOD]</math>
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<math>[COD]=36[AOD]</math>
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<math>\Rightarrow a=4</math>
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Thus we need to find <math>[ABCD]=\frac{79}{2}[AOD]</math>
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Note that <math>\triangle AOD</math> is isosceles with sides <math>4, 4, \frac{2}{3}</math> so we can draw the altitude from D to split it to two right triangles.
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<math>[AOD]=\frac{\sqrt{143}{9}</math>
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Thus <math>[ABCD]=\frac{79\sqrt{143}}{18}\rightarrow\boxed{240}</math>
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==See also==
 
==See also==

Revision as of 12:22, 24 February 2009

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

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

$[AOB]=\frac{[AOD]}{4}$

$[BOC]=\frac{9}{4}[AOD]$

$[COD]=36[AOD]$

$\Rightarrow a=4$

Thus we need to find $[ABCD]=\frac{79}{2}[AOD]$

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.

$[AOD]=\frac{\sqrt{143}{9}$ (Error compiling LaTeX. Unknown error_msg)

Thus $[ABCD]=\frac{79\sqrt{143}}{18}\rightarrow\boxed{240}$


See also


Problem Source

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

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