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Difference between revisions of "Mock AIME 2 2006-2007 Problems/Problem 12"
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== Problem ==
 
== Problem ==
 
In quadrilateral <math>\displaystyle ABCD,</math> <math>\displaystyle m \angle DAC= m\angle DBC </math> and <math>\displaystyle \frac{[ADB]}{[ABC]}=\frac12.</math> If <math>\displaystyle AD=4,</math> <math>\displaystyle BC=6</math>, <math>\displaystyle BO=1,</math> and the area of <math>\displaystyle ABCD</math> is <math>\displaystyle \frac{a\sqrt{b}}{c},</math> where <math>\displaystyle a,b,c</math> are relatively prime positive integers, find <math>\displaystyle a+b+c.</math>
 
In quadrilateral <math>\displaystyle ABCD,</math> <math>\displaystyle m \angle DAC= m\angle DBC </math> and <math>\displaystyle \frac{[ADB]}{[ABC]}=\frac12.</math> If <math>\displaystyle AD=4,</math> <math>\displaystyle BC=6</math>, <math>\displaystyle BO=1,</math> and the area of <math>\displaystyle ABCD</math> is <math>\displaystyle \frac{a\sqrt{b}}{c},</math> where <math>\displaystyle a,b,c</math> are relatively prime positive integers, find <math>\displaystyle a+b+c.</math>
 +
  
 
Note*: <math>\displaystyle[ABC]</math> and <math>\displaystyle[ADB]</math> refer to the areas of triangles <math>\displaystyle ABC</math> and <math>\displaystyle ADB.</math>
 
Note*: <math>\displaystyle[ABC]</math> and <math>\displaystyle[ADB]</math> refer to the areas of triangles <math>\displaystyle ABC</math> and <math>\displaystyle ADB.</math>

Revision as of 18:22, 24 July 2006

Problem

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


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

Problem Source

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

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