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

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== Problem ==
 
== Problem ==
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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>
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In [[quadrilateral]] <math>ABCD,</math> <math>m \angle DAC= m\angle DBC </math> and <math>\frac{[ADB]}{[ABC]}=\frac12.</math> <math>O</math> is defined to be the intersection of the diagonals of <math>ABCD</math>. If <math>AD=4,</math> <math>BC=6</math>, <math>BO=1,</math> and the [[area]] of <math>ABCD</math> is <math>\frac{a\sqrt{b}}{c},</math> where <math>a,b,c</math> are [[relatively prime]] [[positive integer]]s, find <math>a+b+c.</math>
  
  
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Note*: <math>\displaystyle[ABC]</math> and <math>\displaystyle[ADB]</math> refer to the areas of [[triangle]]s <math>\displaystyle ABC</math> and <math>\displaystyle ADB.</math>
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Note*: <math>[ABC]</math> and <math>[ADB]</math> refer to the areas of [[triangle]]s <math>ABC</math> and <math>ADB.</math>
  
 
==Solution==
 
==Solution==
 
{{solution}}
 
{{solution}}
  
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==See also==
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*[[Mock AIME 2 2006-2007/Problem 11 | Previous Problem]]
 
*[[Mock AIME 2 2006-2007/Problem 11 | Previous Problem]]
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*[[Mock AIME 2 2006-2007/Problem 13 | Next Problem]]
 
*[[Mock AIME 2 2006-2007/Problem 13 | Next Problem]]
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*[[Mock AIME 2 2006-2007]]
 
*[[Mock AIME 2 2006-2007]]
  

Revision as of 08:11, 16 August 2008

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

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See also


Problem Source

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

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