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

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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 integer]]s, 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> <math>O</math> is defined to be the intersection of the diagonals of <math>ABCD</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 integer]]s, find <math>\displaystyle a+b+c.</math>




Revision as of 21:09, 16 February 2007

Problem

In quadrilateral $\displaystyle ABCD,$ $\displaystyle m \angle DAC= m\angle DBC$ and $\displaystyle \frac{[ADB]}{[ABC]}=\frac12.$ $O$ is defined to be the intersection of the diagonals of $ABCD$. 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 $\displaystyle ADB.$

Solution

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Problem Source

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

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