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Difference between revisions of "Mock AIME 2 2006-2007 Problems/Problem 9"

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== Problem ==
 
== Problem ==
 
In right triangle <math>\displaystyle ABC,</math> <math>\displaystyle \angle C=90^\circ.</math> Cevians <math>\displaystyle AX</math> and <math>\displaystyle BY</math> intersect at <math>\displaystyle P</math> and are drawn to <math>\displaystyle BC</math> and <math>\displaystyle AC</math> respectively such that <math>\displaystyle \frac{BX}{CX}=\frac23</math> and <math>\displaystyle \frac{AY}{CY}=\sqrt 3.</math> If <math>\displaystyle \tan \angle APB= \frac{a+b\sqrt{c}}{d},</math> where <math>\displaystyle a,b,</math> and <math>\displaystyle d</math> are relatively prime and <math>\displaystyle c</math> has no perfect square divisors excluding <math>\displaystyle 1,</math> find <math>\displaystyle a+b+c+d.</math>
 
In right triangle <math>\displaystyle ABC,</math> <math>\displaystyle \angle C=90^\circ.</math> Cevians <math>\displaystyle AX</math> and <math>\displaystyle BY</math> intersect at <math>\displaystyle P</math> and are drawn to <math>\displaystyle BC</math> and <math>\displaystyle AC</math> respectively such that <math>\displaystyle \frac{BX}{CX}=\frac23</math> and <math>\displaystyle \frac{AY}{CY}=\sqrt 3.</math> If <math>\displaystyle \tan \angle APB= \frac{a+b\sqrt{c}}{d},</math> where <math>\displaystyle a,b,</math> and <math>\displaystyle d</math> are relatively prime and <math>\displaystyle c</math> has no perfect square divisors excluding <math>\displaystyle 1,</math> find <math>\displaystyle a+b+c+d.</math>
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==Solution==
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{{solution}}
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----
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*[[Mock AIME 2 2006-2007/Problem 8 | Previous Problem]]
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*[[Mock AIME 2 2006-2007/Problem 10 | Next Problem]]
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*[[Mock AIME 2 2006-2007]]

Revision as of 18:48, 22 August 2006

Problem

In right triangle $\displaystyle ABC,$ $\displaystyle \angle C=90^\circ.$ Cevians $\displaystyle AX$ and $\displaystyle BY$ intersect at $\displaystyle P$ and are drawn to $\displaystyle BC$ and $\displaystyle AC$ respectively such that $\displaystyle \frac{BX}{CX}=\frac23$ and $\displaystyle \frac{AY}{CY}=\sqrt 3.$ If $\displaystyle \tan \angle APB= \frac{a+b\sqrt{c}}{d},$ where $\displaystyle a,b,$ and $\displaystyle d$ are relatively prime and $\displaystyle c$ has no perfect square divisors excluding $\displaystyle 1,$ find $\displaystyle a+b+c+d.$

Solution

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