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

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<math>\left\lfloor 1000\left(\frac {4}{7}\right)\right\rfloor = \boxed{571}</math>.
 
<math>\left\lfloor 1000\left(\frac {4}{7}\right)\right\rfloor = \boxed{571}</math>.
  
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== Solution 3 ==
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== Solution 3 (Trig identities)==
  
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Let the measure of <math>\angle BAC</math> be <math>\alpha</math> and <math>\overline{AP}=\overline{PQ}=\overline{QB}=\overline{BC}=x</math>. Because <math>\triangle APQ</math> is isosceles, <math>AQ=2x\cos(\alpha)</math>. So, <math>\overline{AB}=x\left(2\cos(\alpha)+1\right)</math>. <math>\triangle{ABC}</math> is isosceles too, so <math>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)</math>.
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Let <math>\angle BAC= 2\theta</math> and <math>AP=PQ=QB=BC=x</math>. <math>\triangle APQ</math> is isosceles, so <math>AQ=2x\cos 2\theta =2x(1-2\sin^2\theta)</math> and <math>AB= AQ+x=x\left(3-4\sin^2\theta\right)</math>. <math>\triangle{ABC}</math> is isosceles too, so <math>x=BC=2AB\sin\theta</math>. Using the expression for <math>AB</math>, we get <cmath>1=2\left(3\sin\theta-4\sin^3\theta\right)=2\sin3\theta</cmath>by the triple angle formula!
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Simplifying, <math>1=4\cos(\alpha)\sin\left(\frac{\alpha}{2}\right)+2\sin\left(\frac{\alpha}{2}\right)</math>. By double angle formula, we know that <math>\cos(\alpha)=1-2\sin^2\left(\frac{\alpha}{2}\right)</math>.
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It follows now that <math>\angle APQ=140^\circ</math>, <math>\angle ACB=80^\circ</math>, giving <math>r=\tfrac{4}{7}</math>.  
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Applying, <math>4\sin\left(\frac{\alpha}{2}\right)-8\sin^3\left(\frac{\alpha}{2}\right)+2\sin\left(\frac{\alpha}{2}\right)=1</math> and <math>2\left(3\sin\left(\frac{\alpha}{2}\right)-4\sin^3\left(\frac{\alpha}{2}\right)\right)=1</math>.
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The expression in the parentheses though is triple angle formula! Hence, <math>\sin\left(\frac{3\alpha}{2}\right)=\frac{1}{2}</math>, <math>\alpha=20^o</math>. It follows now that <math>\angle APQ=140^o</math>, <math>\angle ACB=80^o</math>. Giving <math>r=\frac{4}{7}</math>.
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The answer is <math>\left\lfloor 1000\left(\frac {4}{7}\right)\right\rfloor = \boxed{571}</math>.
−
<math>\left\lfloor 1000\left(\frac {4}{7}\right)\right\rfloor = \boxed{571}</math>.
 
  
 
== See also ==
 
== See also ==

Revision as of 17:47, 25 January 2022

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 $\lfloor 1000r \rfloor$.

Official Solution (MAA)

[asy]defaultpen(fontsize(10)); size(200); pen p=fontsize(8); pair A,B,C,P,Q; B=MP("B",origin,down+left); C=MP("C",20*right,right+down); A=MP("A",extension(B,dir(80),C,C+dir(100)),up); Q=MP("Q",20*dir(80),left); P=MP("P",Q+(20*dir(60)),right); draw(A--B--C--A, black+1);draw(B--P--Q); MP("x",B,20*dir(75),p); MP("x",P,17*dir(245),p); MP("2x",Q,15*dir(70),p); MP("2x",A,15*dir(-90),p); MP("2y",P,2*left,p); MP("3x",P,10*dir(-95),p); MP("x+y",C,5*dir(135),p); MP("y",B,5*dir(40),p); [/asy]

Let $\angle QPB=x^\circ$. Because $\angle AQP$ is exterior to isosceles triangle $PQB$ its measure is $2x$ and $\angle PAQ$ has the same measure. Because $\angle BPC$ is exterior to $\triangle BPA$ its measure is $3x$. Let $\angle PBC = y^\circ$. It follows that $\angle ACB = x+y$ and that $4x+2y=180^\circ$. Two of the angles of triangle $APQ$ have measure $2x$, and thus the measure of $\angle APQ$ is $2y$. It follows that $AQ=2\cdot AP\cdot \sin y$. Because $AB=AC$ and $AP=QB$, it also follows that $AQ=PC$. Now apply the Law of Sines to triangle $PBC$ to find \[\frac{\sin 3x}{BC}=\frac{\sin y}{PC}=\frac{\sin y}{2\cdot AP\cdot \sin y}= \frac{1}{2\cdot BC}\]because $AP=BC$. Hence $\sin 3x = \frac 12$. Since $4x<180$, this implies that $3x=30$, i.e. $x=10$. Thus $y=70$ and \[r=\frac{10+70}{2\cdot 70}=\frac 47,\]which implies that $1000r = 571 + \frac 37$. So the answer is $\boxed{571}$.


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 (Trig identities)

Let $\angle BAC= 2\theta$ and $AP=PQ=QB=BC=x$. $\triangle APQ$ is isosceles, so $AQ=2x\cos 2\theta =2x(1-2\sin^2\theta)$ and $AB= AQ+x=x\left(3-4\sin^2\theta\right)$. $\triangle{ABC}$ is isosceles too, so $x=BC=2AB\sin\theta$. Using the expression for $AB$, we get \[1=2\left(3\sin\theta-4\sin^3\theta\right)=2\sin3\theta\]by the triple angle formula! It follows now that $\angle APQ=140^\circ$, $\angle ACB=80^\circ$, giving $r=\tfrac{4}{7}$.

The answer is $\left\lfloor 1000\left(\frac {4}{7}\right)\right\rfloor = \boxed{571}$.

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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