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Difference between revisions of "2000 AIME I Problems/Problem 14"

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== Solution ==
 
== Solution ==
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{{solution}}
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<center><asy>defaultpen(fontsize(8)); size(200); pair A=20*dir(80)+20*dir(60)+20*dir(100), B=(0,0), C=20*dir(0), P=20*dir(80)+20*dir(60), Q=20*dir(80), R=20*dir(60), S; S=intersectionpoint(Q--C,P--B); draw(A--B--C--A);draw(B--P--Q--C--R--Q);draw(A--R--B);draw(P--R--S); label("A",A,(0,1));label("B",B,(-1,-1));label("C",C,(1,-1));label("P",P,(1,1)); label("Q",Q,(-1,1));label("R",R,(1,0));label("S",S,(-1,0));
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</asy></center>Let point <math>R</math> be in <math>\triangle ABC</math> such that <math>QB = BR = RP</math>, thus <math>PQBR</math> is a rhombus and <math>APRB</math> is an isosceles trapezoid.
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Let <math>\angle BAC = \angle QBR = \angle QPR = 2x</math> and <math>\angle ABC = \angle ACB = y</math>, which means <math>x + y = 90</math>.
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<math>\triangle QBC</math> is isosceles with <math>QB = BC</math>, so <math>\angle BCQ = 90 - \frac {y}{2}</math>.
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Let <math>S</math> be the intersection of <math>QC</math>and <math>BR</math>. Since <math>\angle BCQ = \angle BQC = \angle BRS</math>, <math>BCRS</math> is cyclic, which means <math>\angle RBS = \angle RCS = x</math>.
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Since <math>APRB</math> is an isosceles trapezoid, <math>BP = AR</math>, but since <math>AR</math> bisects <math>\angle BAC</math>, <math>\angle ABR = \angle ACR = 2x</math>.
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Therefore we have that <math>\angle ACB = \angle ACR + \angle RCS + \angle QCB = 2x + x + 90 - \frac {y}{2} = y</math>.
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We solve the simultaneous equations <math>x + y = 90</math> and <math>2x + x + 90 - \frac {y}{2} = y</math> to get <math>x = 10</math> and <math>y = 80</math>.
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<math>\angle APQ = 180 - 4x = 140</math>, <math>\angle ACB = 80</math>, so <math>r = \frac {80}{140} = \frac {4}{7}</math>.
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<math>\left\lfloor 1000\left(\frac {4}{7}\right)\right\rfloor = \boxed{571}</math>
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2000|n=I|num-b=13|num-a=15}}
 
{{AIME box|year=2000|n=I|num-b=13|num-a=15}}

Revision as of 17:22, 3 February 2008

Problem

In triangle $ABC,$ it is given that angles $B$ and $C$ are congruent. Points $P$ and $Q$ lie on $\overline{AC}$ and $\overline{AB},$ respectively, so that $AP = PQ = QB = BC.$ Angle $ACB$ is $r$ times as large as angle $APQ,$ where $r$ is a positive real number. Find the greatest integer that does not exceed $1000r$.

Solution

[asy]defaultpen(fontsize(8)); size(200); pair A=20*dir(80)+20*dir(60)+20*dir(100), B=(0,0), C=20*dir(0), P=20*dir(80)+20*dir(60), Q=20*dir(80), R=20*dir(60), S; S=intersectionpoint(Q--C,P--B); draw(A--B--C--A);draw(B--P--Q--C--R--Q);draw(A--R--B);draw(P--R--S); label("A",A,(0,1));label("B",B,(-1,-1));label("C",C,(1,-1));label("P",P,(1,1)); label("Q",Q,(-1,1));label("R",R,(1,0));label("S",S,(-1,0)); [/asy]

Let point $R$ be in $\triangle ABC$ such that $QB = BR = RP$, thus $PQBR$ is a rhombus and $APRB$ is an isosceles trapezoid.

Let $\angle BAC = \angle QBR = \angle QPR = 2x$ and $\angle ABC = \angle ACB = y$, which means $x + y = 90$. $\triangle QBC$ is isosceles with $QB = BC$, so $\angle BCQ = 90 - \frac {y}{2}$. Let $S$ be the intersection of $QC$and $BR$. Since $\angle BCQ = \angle BQC = \angle BRS$, $BCRS$ is cyclic, which means $\angle RBS = \angle RCS = x$. Since $APRB$ is an isosceles trapezoid, $BP = AR$, but since $AR$ bisects $\angle BAC$, $\angle ABR = \angle ACR = 2x$. Therefore we have that $\angle ACB = \angle ACR + \angle RCS + \angle QCB = 2x + x + 90 - \frac {y}{2} = y$. We solve the simultaneous equations $x + y = 90$ and $2x + x + 90 - \frac {y}{2} = y$ to get $x = 10$ and $y = 80$. $\angle APQ = 180 - 4x = 140$, $\angle ACB = 80$, so $r = \frac {80}{140} = \frac {4}{7}$. $\left\lfloor 1000\left(\frac {4}{7}\right)\right\rfloor = \boxed{571}$

See also

2000 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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