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Difference between revisions of "Mock AIME 2 2006-2007 Problems/Problem 9"
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== Problem ==
 
== Problem ==
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In right triangle <math>ABC,</math> <math>\angle C=90^\circ.</math> Cevians <math>AX</math> and <math>BY</math> intersect at <math>P</math> and are drawn to <math>BC</math> and <math>AC</math> respectively such that <math>\frac{BX}{CX}=\frac23</math> and <math>\frac{AY}{CY}=\sqrt 3.</math> If <math>\tan \angle APB= \frac{a+b\sqrt{c}}{d},</math> where <math>a,b,</math> and <math>d</math> are relatively prime and <math>c</math> has no perfect square divisors excluding <math>1,</math> find <math>a+b+c+d.</math>
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In right triangle <math>ABC,</math> <math>\angle C=90^\circ.</math> Cevians <math>AX</math> and <math>BY</math> intersect at <math>P</math> and are drawn to <math>BC</math> and <math>AC</math> respectively such that <math>\frac{BX}{CX}=\frac23</math>, <math>\frac{AY}{CY}=\sqrt 3,</math> and <math>CY=CX-BX</math>. If <math>\tan \angle APB= -\frac{a+b\sqrt{c}}{d},</math> where <math>a,b,</math> and <math>d</math> are relatively prime and <math>c</math> has no perfect square divisors excluding <math>1,</math> find <math>a+b+c+d.</math>
  
 
==Solution==
 
==Solution==
−
{{solution}}
+
<asy>
 +
import olympiad;
 +
size(200);
 +
defaultpen(linewidth(0.8));
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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);
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label("$A$",A,N);
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label("$B$",B,E);
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label("$C$",C,SW);
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label("$X$",X,S);
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label("$Y$",Y,W);
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pair P=extension(A,X,B,Y);
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pair D=foot(P,B,C),E=foot(P,A,C);
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draw(D--P--E,linetype("4 4"));
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draw(rightanglemark(A,C,B,5)^^rightanglemark(A,E,P,5)^^rightanglemark(P,D,B,5));
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label("$D$",D,S);
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label("$E$",E,W);
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label("$P$",P,0.5(dir(P--B)+dir(P--A)));
 +
</asy>
 +
 
 +
Define <math>CX=3x</math>, <math>XB=2x</math>, <math>CY=y</math>, and <math>YA=y\sqrt3</math>.  Note that <math>CY=CX-BX</math> implies <math>y=3x-2x=x</math>.
 +
 
 +
Let <math>D</math> and <math>E</math> be the projections of <math>P</math> onto the legs <math>AC</math> and <math>BC</math> respectively.  Remark that <math>\tan\angle DPB=\tan \angle CXB=5</math> and <math>\tan\angle APE=\tan\angle AXC=1+\sqrt3</math>, so <cmath>\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*}</cmath>  Since <math>\angle APB+(\angle APE+\angle DPB)=270^\circ</math>, we have <cmath>\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*}</cmath> The requested answer is thus <math>9+26+3+33=\boxed{071}.</math>
  
 
==See Also==
 
==See Also==
 
{{Mock AIME box|year=2006-2007|n=2|num-b=8|num-a=10}}
 
{{Mock AIME box|year=2006-2007|n=2|num-b=8|num-a=10}}

Latest revision as of 22:22, 3 May 2014

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

[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 $9+26+3+33=\boxed{071}.$

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
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