2000 AIME I Problems/Problem 14

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 $r = \frac {80}{140} = \frac {4}{7}$ . $\left\lfloor 1000\left(\frac {4}{7}\right)\right\rfloor = \boxed{571}$

2000 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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2000 AIME I Problems/Problem 14

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