Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(easier solution, on same principles)
m (typo fixes)
Line 9: Line 9:
 
Let point <math>R</math> be in <math>\triangle ABC</math> such that <math>QB = BR = RP</math>. Then <math>PQBR</math> is a [[rhombus]], so <math>AB \parallel PR</math> and <math>APRB</math> is an [[isosceles trapezoid]]. Since <math>\overline{PB}</math> bisects <math>\angle QBR</math>, it follows by symmetry in trapezoid <math>APRB</math> that <math>\overline{RA}</math> bisects <math>\angle BAC</math>. Thus <math>R</math> lies on the perpendicular bisector of <math>\overline{BC}</math>, and <math>BC = BR = RC</math>. Hence <math>\triangle BCR</math> is an [[equilateral triangle]].
 
Let point <math>R</math> be in <math>\triangle ABC</math> such that <math>QB = BR = RP</math>. Then <math>PQBR</math> is a [[rhombus]], so <math>AB \parallel PR</math> and <math>APRB</math> is an [[isosceles trapezoid]]. Since <math>\overline{PB}</math> bisects <math>\angle QBR</math>, it follows by symmetry in trapezoid <math>APRB</math> that <math>\overline{RA}</math> bisects <math>\angle BAC</math>. Thus <math>R</math> lies on the perpendicular bisector of <math>\overline{BC}</math>, and <math>BC = BR = RC</math>. Hence <math>\triangle BCR</math> is an [[equilateral triangle]].
  
−
Now <math>\angle ABR = \angle BAC = \angle ACR</math>, and the sum of the angles in <math>\triangle ABC</math> is <math>3\angle BAC + 2 \cdot 60^{\circ} = 180^{\circ} \Longrightarrow \angle BAC = 20^{\circ}</math>. Then <math>\angle APQ = 140^{\circ}</math> and <math>\angle ACB = 80^{\circ}</math>, so the answer is <math>\left\lfloor 1000 \cdot \frac{80}{140} \right\rfloor = \left\lfloor \frac{4000}{7} \right\rfloor = \boxed{571}</math>.
+
Now <math>\angle ABR = \angle BAC = \angle ACR</math>, and the sum of the angles in <math>\triangle ABC</math> is <math>\angle ABR + 60^{\circ} + \angle BAC + \angle ACR + 60^{\circ} = 3\angle BAC + 120^{\circ} = 180^{\circ} \Longrightarrow \angle BAC = 20^{\circ}</math>. Then <math>\angle APQ = 140^{\circ}</math> and <math>\angle ACB = 80^{\circ}</math>, so the answer is <math>\left\lfloor 1000 \cdot \frac{80}{140} \right\rfloor = \left\lfloor \frac{4000}{7} \right\rfloor = \boxed{571}</math>.
  
 
=== Solution 2 ===
 
=== Solution 2 ===
Line 18: Line 18:
 
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>.  
 
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>.  
 
<math>\triangle QBC</math> is isosceles with <math>QB = BC</math>, so <math>\angle BCQ = 90 - \frac {y}{2}</math>.
 
<math>\triangle QBC</math> is isosceles with <math>QB = BC</math>, so <math>\angle BCQ = 90 - \frac {y}{2}</math>.
−
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 quadrilateral|cyclic]], which means <math>\angle RBS = \angle RCS = x</math>.
+
Let <math>S</math> be the intersection of <math>QC</math> and <math>BP</math>. Since <math>\angle BCQ = \angle BQC = \angle BRS</math>, <math>BCRS</math> is [[cyclic quadrilateral|cyclic]], which means <math>\angle RBS = \angle RCS = x</math>.
 
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>.   
 
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>.   
  

Revision as of 16:36, 14 August 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

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

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_I_Problems/Problem_14&oldid=27416"