Difference between revisions of "2000 AIME I Problems/Problem 14"
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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>. | ||
+ | |||
+ | Note: We can also solve this quickly through angle chasing. If <math>\angle ABC = \theta</math>, <math>\angle ABR = \theta - 60</math>. Since <math>QPRB</math> is a rhombus, <math>\angle PQB = 180-(\theta-60) = 240 - \theta</math> so <math>\angle AQP = \theta - 60 = \angle PAQ</math>. Thus, we have that <math>\theta + \theta + \theta - 60 = 180 \implies \theta = 80</math>. The rest of the problem is easily solved from there (<math>\angle QPA = 140</math> so <math>\left\lfloor 1000 \cdot \frac{80}{140} \right\rfloor = \left\lfloor 1000 \cdot \frac{4}{7} \right\rfloor = 571</math>. | ||
=== Solution 2 === | === Solution 2 === |
Revision as of 22:21, 29 November 2021
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
.
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
.
Note: We can also solve this quickly through angle chasing. If ,
. Since
is a rhombus,
so
. Thus, we have that
. The rest of the problem is easily solved from there (
so
.
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
Let the measure of be
and
. Because
is isosceles,
. So,
.
is isosceles too, so
.
Simplifying,
. By double angle formula, we know that
.
Applying,
and
.
The expression in the parentheses though is triple angle formula! Hence,
,
. It follows now that
,
. Giving
.
.
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.