Difference between revisions of "2006 AIME II Problems/Problem 12"
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Lines <math>l_1</math> and <math>l_2</math> are constructed such that <math>AEFD</math> is a parallelogram, hence <math>DF = 13</math>. Since <math>BAC</math> is equilateral with angle of <math>60^{\circ}</math>, angle <math>D</math> is <math>120^{\circ}</math>. Use law of cosines to find <math>AF = \sqrt{433}</math>. Then use law of sines to find angle <math>BAG</math> and <math>GAC</math>. Next we use Ptolemy's Theorem on <math>ABGC</math> to find that <math>CG + BG = AG</math>. Next we use law of cosine on triangles <math>BAG</math> and <math>GAC</math>, solving for BG and CG respectively. Subtract the two equations and divide out a <math>BG + CG</math> to find the value of <math>CG - BG</math>. Next, <math>AG = 2\cdot R \cos{\theta}</math>, where R is radius of circle <math> = 2</math> and <math>\theta = </math> angle <math>BAG</math>. We already know sine of the angle so find cosine, hence we have found <math>AG</math>. At this point it is system of equation yielding <math>CG = \frac{26\sqrt{3}}{\sqrt{433}}</math> and <math>BG = \frac{22\sqrt{3}}{\sqrt{433}}</math>. Given <math>[CBG] = \frac{BC \cdot CG \cdot BG}{4R}</math>, and <math>BC = 2\sqrt{3}</math> by <math>30-60-90</math> triangle, we can evaluate to find <math>[CBG] = \frac{429\sqrt{3}}{433}</math>, to give answer = <math>\boxed{865}</math>. | Lines <math>l_1</math> and <math>l_2</math> are constructed such that <math>AEFD</math> is a parallelogram, hence <math>DF = 13</math>. Since <math>BAC</math> is equilateral with angle of <math>60^{\circ}</math>, angle <math>D</math> is <math>120^{\circ}</math>. Use law of cosines to find <math>AF = \sqrt{433}</math>. Then use law of sines to find angle <math>BAG</math> and <math>GAC</math>. Next we use Ptolemy's Theorem on <math>ABGC</math> to find that <math>CG + BG = AG</math>. Next we use law of cosine on triangles <math>BAG</math> and <math>GAC</math>, solving for BG and CG respectively. Subtract the two equations and divide out a <math>BG + CG</math> to find the value of <math>CG - BG</math>. Next, <math>AG = 2\cdot R \cos{\theta}</math>, where R is radius of circle <math> = 2</math> and <math>\theta = </math> angle <math>BAG</math>. We already know sine of the angle so find cosine, hence we have found <math>AG</math>. At this point it is system of equation yielding <math>CG = \frac{26\sqrt{3}}{\sqrt{433}}</math> and <math>BG = \frac{22\sqrt{3}}{\sqrt{433}}</math>. Given <math>[CBG] = \frac{BC \cdot CG \cdot BG}{4R}</math>, and <math>BC = 2\sqrt{3}</math> by <math>30-60-90</math> triangle, we can evaluate to find <math>[CBG] = \frac{429\sqrt{3}}{433}</math>, to give answer = <math>\boxed{865}</math>. | ||
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== See also == | == See also == |
Revision as of 15:43, 6 December 2020
Contents
Problem
Equilateral is inscribed in a circle of radius
. Extend
through
to point
so that
and extend
through
to point
so that
Through
draw a line
parallel to
and through
draw a line
parallel to
Let
be the intersection of
and
Let
be the point on the circle that is collinear with
and
and distinct from
Given that the area of
can be expressed in the form
where
and
are positive integers,
and
are relatively prime, and
is not divisible by the square of any prime, find
Solution 1
![[asy] size(250); pointpen = black; pathpen = black + linewidth(0.65); pen s = fontsize(8); pair A=(0,0),B=(-3^.5,-3),C=(3^.5,-3),D=13*expi(-2*pi/3),E1=11*expi(-pi/3),F=E1+D; path O = CP((0,-2),A); pair G = OP(A--F,O); D(MP("A",A,N,s)--MP("B",B,W,s)--MP("C",C,E,s)--cycle);D(O); D(B--MP("D",D,W,s)--MP("F",F,s)--MP("E",E1,E,s)--C); D(A--F);D(B--MP("G",G,SW,s)--C); MP("11",(A+E1)/2,NE);MP("13",(A+D)/2,NW);MP("l_1",(D+F)/2,SW);MP("l_2",(E1+F)/2,SE); [/asy]](http://latex.artofproblemsolving.com/2/9/9/299c51b0ad858e4078795c0f549803abc5d0ec0d.png)
Notice that because
. Also,
because they both correspond to arc
. So
.
Because the ratio of the area of two similar figures is the square of the ratio of the corresponding sides, . Therefore, the answer is
.
Solution 2: Analytic Geometry/Cord Bash
Solution by e_power_pi_times_i/edited by srisainandan6
Let the center of the circle be and the origin. Then,
,
,
.
and
can be calculated easily knowing
and
,
,
. As
and
are parallel to
and
,
.
and
is the intersection between
and circle
. Therefore
. Using the Shoelace Theorem,
, so the answer is
. Note that although the solution may appear short, actually getting all the coordinates take a while as there is plenty of computation
Solution 3: Trig
Lines and
are constructed such that
is a parallelogram, hence
. Since
is equilateral with angle of
, angle
is
. Use law of cosines to find
. Then use law of sines to find angle
and
. Next we use Ptolemy's Theorem on
to find that
. Next we use law of cosine on triangles
and
, solving for BG and CG respectively. Subtract the two equations and divide out a
to find the value of
. Next,
, where R is radius of circle
and
angle
. We already know sine of the angle so find cosine, hence we have found
. At this point it is system of equation yielding
and
. Given
, and
by
triangle, we can evaluate to find
, to give answer =
.
See also
2006 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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