Difference between revisions of "Mock AIME 2 2006-2007 Problems/Problem 9"
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==Solution== | ==Solution== | ||
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<asy> | <asy> | ||
import olympiad; | import olympiad; | ||
Latest revision as of 22:22, 3 May 2014
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 
Solution
![[asy] import olympiad; size(200); defaultpen(linewidth(0.8)); pair A=(0,6),B=(5,0),C=origin,X=(3,0),Y=A/(sqrt(3)+1); draw(A--B--C--A--X^^Y--B); label("$A$",A,N); label("$B$",B,E); label("$C$",C,SW); label("$X$",X,S); label("$Y$",Y,W); pair P=extension(A,X,B,Y); pair D=foot(P,B,C),E=foot(P,A,C); draw(D--P--E,linetype("4 4")); draw(rightanglemark(A,C,B,5)^^rightanglemark(A,E,P,5)^^rightanglemark(P,D,B,5)); label("$D$",D,S); label("$E$",E,W); label("$P$",P,0.5(dir(P--B)+dir(P--A))); [/asy]](http://latex.artofproblemsolving.com/1/9/2/192939ac7d2bcccc913e3eda6ab3431d13ed2652.png)
Define 
, 
, 
, and 
. Note that implies 
.
Let 
and 
be the projections of onto the legs and respectively. Remark that 
and 
, so 
Since 
, we have ![\begin{align*}\tan\angle APB&=\tan[270^\circ-(\angle APE+\angle BPD)]\\&=\cot (\angle APE+\angle BPD)\\&=-\dfrac{4+5\sqrt3}{6+\sqrt3}=-\dfrac{9+26\sqrt3}{33}.\end{align*}](http://latex.artofproblemsolving.com/c/f/1/cf1419a232c670d6a0db5e9fc699d07973f47a60.png)
The requested answer is thus 
See Also
| Mock AIME 2 2006-2007 (Problems, Source) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
