Mock AIME 2 2006-2007 Problems/Problem 9

Problem

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

Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it. $[asy] import olympiad; size(200); defaultpen(linewidth(0.8)); pair A=(0,6),B=(5,0),C=origin,X=(3,0),Y=A/(sqrt(3)+1); draw(A--B--C--A--X^^Y--B); label("$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$",P,0.5(dir(P--B)+dir(P--A))); [/asy]$

Define $CX=3x$ , $XB=2x$ , $CY=y$ , and $YA=y\sqrt3$ . Note that $CY=CX-BX$ implies $y=3x-2x=x$ .

Let $D$ and $E$ be the projections of $P$ onto the legs $AC$ and $BC$ respectively. Remark that $\tan\angle DPB=\tan \angle CXB=5$ and $\tan\angle APE=\tan\angle AXC=1+\sqrt3$ , so $\begin{align*}\tan(\angle DPB+\angle APE)=\dfrac{\tan\angle DPB+\tan\angle APE}{1-\tan\angle DPB\tan\angle APE}=\dfrac{5+1+\sqrt3}{1-5(1+\sqrt3)}=-\dfrac{6+\sqrt3}{4+5\sqrt3}.\end{align*}$ Since $\angle APB+(\angle APE+\angle DPB)=270^\circ$ , we have $\begin{align*}\tan\angle APB&=\tan[270^\circ-(\angle APE+\angle BPD)]\\&=\cot (\angle APE+\angle BPD)\\&=-\dfrac{4+5\sqrt3}{6+\sqrt3}=-\dfrac{9+26\sqrt3}{33}.\end{align*}$ The requested answer is thus $9+26+3+33=\boxed{071}.$

Mock AIME 2 2006-2007 (Problems, Source)
Preceded by Problem 8	Followed by Problem 10
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Mock AIME 2 2006-2007 Problems/Problem 9

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