Difference between revisions of "2000 AIME I Problems/Problem 14"
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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> | + | 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 === | ||
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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>. | 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> | + | 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 it is given that angles
and
are congruent. Points
and
lie on
and
respectively, so that
Angle
is
times as large as angle
where
is a positive real number. Find the greatest integer that does not exceed
.
Contents
[hide]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]](http://latex.artofproblemsolving.com/f/1/f/f1f597c511a21d50292f1daf4ee93815c1404c82.png)
Let point be in
such that
. Then
is a rhombus, so
and
is an isosceles trapezoid. Since
bisects
, it follows by symmetry in trapezoid
that
bisects
. Thus
lies on the perpendicular bisector of
, and
. Hence
is an equilateral triangle.
Now , and the sum of the angles in
is
. Then
and
, so the answer is
.
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]](http://latex.artofproblemsolving.com/1/6/6/166838a1bc564730cc23500919254edab0c132e2.png)
Again, construct as above.
Let and
, which means
.
is isosceles with
, so
.
Let
be the intersection of
and
. Since
,
is cyclic, which means
.
Since
is an isosceles trapezoid,
, but since
bisects
,
.
Therefore we have that .
We solve the simultaneous equations
and
to get
and
.
,
, so
.
.
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 |