# 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

### Solution 1

$[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); draw(A--B--C--A);draw(P--Q);draw(A--R--B);draw(P--R);D(R--C,dashed); 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)); [/asy]$

Let point $R$ be in $\triangle ABC$ such that $QB = BR = RP$. Then $PQBR$ is a rhombus, so $AB \parallel PR$ and $APRB$ is an isosceles trapezoid. Since $\overline{PB}$ bisects $\angle QBR$, it follows by symmetry in trapezoid $APRB$ that $\overline{RA}$ bisects $\angle BAC$. Thus $R$ lies on the perpendicular bisector of $\overline{BC}$, and $BC = BR = RC$. Hence $\triangle BCR$ is an equilateral triangle.

Now $\angle ABR = \angle BAC = \angle ACR$, and the sum of the angles in $\triangle ABC$ is $\angle ABR + 60^{\circ} + \angle BAC + \angle ACR + 60^{\circ} = 3\angle BAC + 120^{\circ} = 180^{\circ} \Longrightarrow \angle BAC = 20^{\circ}$. Then $\angle APQ = 140^{\circ}$ and $\angle ACB = 80^{\circ}$, so the answer is $\left\lfloor 1000 \cdot \frac{80}{140} \right\rfloor = \left\lfloor \frac{4000}{7} \right\rfloor = \boxed{571}$.

### Solution 2

$[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]$

Again, construct $R$ as above.

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

### Solution 3

Let the measure of $\angle BAC$ be $\alpha$ and $\overline{AP}=\overline{PQ}=\overline{QB}=\overline{BC}=x$. Because $\triangle APQ$ is isosceles, $AQ=2x\cos(\alpha)$. So, $\overline{AB}=x\left(2\cos(\alpha)+1\right)$. $\triangle{ABC}$ is isosceles too, so $x=\overline{BC}=2\overline{AC}\sin\left(\frac{\alpha}{2}\right)=2x\left(2\cos(\alpha)+1\right)\sin\left(\frac{\alpha}{2}\right)$. Simplifying, $1=4\cos(\alpha)\sin\left(\frac{\alpha}{2}\right)+2sin\left(\frac{\alpha}{2}\right)$. By double angle formula, we know that $\cos(\alpha)=1-2\sin^2\left(\frac{\alpha}{2}\right)$. Applying, $4\sin\left(\frac{\alpha}{2}\right)-8\sin^3\left(\frac{\alpha}{2}\right)+2\sin\left(\frac{\alpha}{2}\right)=1$ and $2\left(3\sin\left(\frac{\alpha}{2}\right)-4\sin^3\left(\frac{\alpha}{2}\right)\right)=1$. The expression in the parentheses though is triple angle formula! Hence, $\sin\left(\frac{3\alpha}{2}\right)=\frac{1}{2}$, $\alpha=20^o$. It follows now that $\angle APQ=140^o$, $\angle ACB=80^o$. Giving $r=\frac{4}{7}$. $\left\lfloor 1000\left(\frac {4}{7}\right)\right\rfloor = \boxed{571}$.