Difference between revisions of "Mock AIME 2 2006-2007 Problems/Problem 9"
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− | In right triangle <math>\displaystyle ABC,</math> <math>\displaystyle \angle C=90^\circ.</math> Cevians <math>\displaystyle AX</math> and <math>\displaystyle BY</math> 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> |
Revision as of 13:05, 25 July 2006
Problem
In right triangle
Cevians
and
intersect at
and are drawn to
and
respectively such that
and
If
where
and
are relatively prime and
has no perfect square divisors excluding
find