Difference between revisions of "2000 AIME I Problems/Problem 14"
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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); | 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> | </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 = \ | + | 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>. |
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== Solution 1 == | == Solution 1 == |
Revision as of 17:50, 25 January 2022
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
![[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 3 (Trig identities)
Let and
.
is isosceles, so
and
.
is isosceles too, so
. Using the expression for
, we get
by the triple angle formula!
It follows now that
,
, giving
.
The answer is .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.