Difference between revisions of "2000 AIME I Problems/Problem 14"
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== Problem == | == Problem == | ||
− | In triangle <math>ABC,</math> it is given that angles <math>B</math> and <math>C</math> are [[congruent (geometry)|congruent]]. Points <math>P</math> and <math>Q</math> lie on <math>\overline{AC}</math> and <math>\overline{AB},</math> respectively, so that <math>AP = PQ = QB = BC.</math> Angle <math>ACB</math> is <math>r</math> times as large as angle <math>APQ,</math> where <math>r</math> is a positive real number. Find | + | In triangle <math>ABC,</math> it is given that angles <math>B</math> and <math>C</math> are [[congruent (geometry)|congruent]]. Points <math>P</math> and <math>Q</math> lie on <math>\overline{AC}</math> and <math>\overline{AB},</math> respectively, so that <math>AP = PQ = QB = BC.</math> Angle <math>ACB</math> is <math>r</math> times as large as angle <math>APQ,</math> where <math>r</math> is a positive real number. Find <math>\lfloor 1000r \rfloor</math>. |
__TOC__ | __TOC__ | ||
− | == Solution == | + | == Official Solution (MAA)== |
− | === Solution 1 | + | |
+ | <center><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></center> | ||
+ | Let <math>\angle QPB=x^\circ</math>. Because <math>\angle AQP</math> is exterior to isosceles triangle <math>PQB</math> its measure is <math>2x</math> and <math>\angle PAQ</math> has the same measure. Because <math>\angle BPC</math> is exterior to <math>\triangle BPA</math> its measure is <math>3x</math>. Let <math>\angle PBC = y^\circ</math>. It follows that <math>\angle ACB = x+y</math> and that <math>4x+2y=180^\circ</math>. Two of the angles of triangle <math>APQ</math> have measure <math>2x</math>, and thus the measure of <math>\angle APQ</math> is <math>2y</math>. It follows that <math>AQ=2\cdot AP\cdot \sin y</math>. Because <math>AB=AC</math> and <math>AP=QB</math>, it also follows that <math>AQ=PC</math>. Now apply the Law of Sines to triangle <math>PBC</math> to find <cmath>\frac{\sin 3x}{BC}=\frac{\sin y}{PC}=\frac{\sin y}{2\cdot AP\cdot \sin y}= \frac{1}{2\cdot BC}</cmath>because <math>AP=BC</math>. Hence <math>\sin 3x = \tfrac 12</math>. Since <math>4x<180</math>, this implies that <math>3x=30</math>, i.e. <math>x=10</math>. Thus <math>y=70</math> and <cmath>r=\frac{10+70}{2\cdot 70}=\frac 47,</cmath>which implies that <math>1000r = 571 + \tfrac 37</math>. So the answer is <math>\boxed{571}</math>. | ||
+ | |||
+ | == Solution 1 == | ||
<center><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(" | <center><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(" | ||
</asy></center> | </asy></center> | ||
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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>. | 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 (Law of sines) == |
+ | |||
+ | Let <math>AP=PQ=QB=BC=x</math> and <math>A</math> be the measure of <math>\angle BAC</math>. Since <math>\triangle APQ</math> and <math>\triangle ABC</math> are isoceles, <math>\angle APQ = 180-2A</math> and <math>\angle ACB = 90-\frac{A}{2}</math>. | ||
+ | Because <math>\triangle APQ</math> and <math>\triangle ABC</math> both have a side length <math>x</math> opposite <math>\angle BAC</math>, by the law of sines: | ||
+ | |||
+ | <math>\frac{x}{\sin A}=\frac{AQ}{\sin(180-2A)}=\frac{AQ+x}{\sin(90-\frac{A}{2})}</math> | ||
+ | |||
+ | Simplifying, this becomes | ||
+ | |||
+ | <math>\frac{x}{\sin A}=\frac{AQ}{\sin 2A}=\frac{AQ+x}{\cos \frac{A}{2}}</math> | ||
+ | |||
+ | From the first two fractions, | ||
+ | |||
+ | <math>AQ\cdot \sin A = x \cdot \sin 2A = x \cdot (2\sin A \cos A) \Longrightarrow AQ=x\cdot 2\cos A</math> | ||
+ | |||
+ | Substituting, we have from the first and third fractions, | ||
+ | |||
+ | <math>\frac{x}{\sin A}=\frac{x\cdot 2\cos A + x}{\cos \frac{A}{2}} \Longrightarrow 2\cos A\sin A + \sin A=\sin 2A + \sin A = \cos \frac{A}{2}</math> | ||
+ | |||
+ | By sum-to-product, | ||
+ | |||
+ | <math>\sin 2A + \sin A = 2\sin \frac{3A}{2} \cos \frac{A}{2}</math> | ||
+ | |||
+ | Thus, | ||
+ | |||
+ | <math>2\sin \frac{3A}{2} \cos \frac{A}{2} = \cos \frac{A}{2} \Longrightarrow \sin \frac{3A}{2} = \frac{1}{2}</math> | ||
+ | |||
+ | Because <math>BC=QB<AB</math>, <math>\angle A</math> is acute, so <math>\frac{3A}{2}=30 \Longrightarrow A=20</math> | ||
+ | |||
+ | <math>\angle ACB = \frac{180-20}{2}=80</math>, <math>\angle APQ = 180-2\cdot 20 = 140 \Longrightarrow r=\frac{4}{7}</math> | ||
+ | |||
+ | <math>1000r=\frac{4000}{7}=\boxed{571}.\overline{428571}</math> | ||
+ | |||
+ | ~bad_at_mathcounts | ||
+ | |||
+ | |||
+ | == Solution 3 == | ||
<center><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(" | <center><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(" | ||
</asy></center> | </asy></center> | ||
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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>. | ||
− | === | + | == Solution 4 (Trig identities)== |
+ | |||
+ | 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! Thus <math>\theta=10^\circ</math> and <math>\angle A = 2\theta=20^\circ</math>. | ||
+ | It follows now that <math>\angle APQ=140^\circ</math>, <math>\angle ACB=80^\circ</math>, giving <math>r=\tfrac{4}{7}</math>, which implies that <math>1000r = 571 + \tfrac 37</math>. So the answer is <math>\boxed{571}</math>. | ||
− | + | ==Solution 5== | |
− | + | <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(0)+20*dir(80), D=20*dir(0)+20*dir(80)+20*dir(60)+20*dir(100); draw(R--A--B--C--D--A--C);draw(Q--P--R--Q--C); draw(B--P--D); label("A",A,NW); label("B",B,SW); label("C",C,SE); label("D",D,NE); label("P",P,W); label("Q",Q,W); label("R",R,E);</asy> | |
− | + | Reflect <math>\triangle ABC</math> over <math>BC</math> and translate it to attach side <math>AB</math> onto <math>AC</math>, mapping <math>\triangle ABC</math> to <math>\triangle CAD</math>. Point <math>P</math> maps to <math>R</math>, and <math>Q</math> maps to <math>P</math>. Then we have that <math>BC=BQ=QP=PA=AD=PR=RC</math>. Notice how <math>BQ=RC</math> and <math>BQ\parallel RC</math>, so <math>BQRC</math> is a parallelogram and <math>QR=BC</math>. But <math>BC=QP=PR</math>, so <math>\triangle QPR</math> is actually equilateral. Set <math>\angle BAC=\angle ACD=x</math>. Then notice that <math>\angle QPC=\angle PQA+\angle PAQ=2x</math>, but <math>\angle RPC=\angle PQA=x</math>. Thus <math>\angle QPR=3x=60</math>, so <math>x=20</math>. Thus <math>\angle QPA=140^{\circ}</math> and <math>\angle BCA=80^{\circ}</math>, so <math>r=\frac{80}{140}=\frac{4}{7}</math>. The answer is <math>\left \lfloor \frac{4000}{7}\right \rfloor =\boxed{571}</math>. | |
− | + | ||
− | <math>\left\lfloor | + | ~ethanzhang1001 |
== See also == | == See also == |
Latest revision as of 22:17, 31 December 2024
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
.
Contents
[hide]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]](http://latex.artofproblemsolving.com/5/4/e/54e1500cdc53113031fabaa71353653e3e4801a8.png)
Let . Because
is exterior to isosceles triangle
its measure is
and
has the same measure. Because
is exterior to
its measure is
. Let
. It follows that
and that
. Two of the angles of triangle
have measure
, and thus the measure of
is
. It follows that
. Because
and
, it also follows that
. Now apply the Law of Sines to triangle
to find
because
. Hence
. Since
, this implies that
, i.e.
. Thus
and
which implies that
. So the answer is
.
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 (Law of sines)
Let and
be the measure of
. Since
and
are isoceles,
and
.
Because
and
both have a side length
opposite
, by the law of sines:
Simplifying, this becomes
From the first two fractions,
Substituting, we have from the first and third fractions,
By sum-to-product,
Thus,
Because ,
is acute, so
,
~bad_at_mathcounts
Solution 3
![[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
.
.
Solution 4 (Trig identities)
Let and
.
is isosceles, so
and
.
is isosceles too, so
. Using the expression for
, we get
by the triple angle formula! Thus
and
.
It follows now that
,
, giving
, which implies that
. So the answer is
.
Solution 5
Reflect
over
and translate it to attach side
onto
, mapping
to
. Point
maps to
, and
maps to
. Then we have that
. Notice how
and
, so
is a parallelogram and
. But
, so
is actually equilateral. Set
. Then notice that
, but
. Thus
, so
. Thus
and
, so
. The answer is
.
~ethanzhang1001
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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