Difference between revisions of "2006 AIME II Problems/Problem 12"
m (→Problem) |
|||
(18 intermediate revisions by 11 users not shown) | |||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
− | [[Equilateral triangle | Equilateral]] <math> \triangle ABC </math> is inscribed in a [[circle]] of [[radius]] 2. Extend <math> \overline{AB} </math> through <math> B </math> to point <math> D </math> so that <math> AD=13, </math> and extend <math> \overline{AC} </math> through <math> C </math> to point <math> E </math> so that <math> AE = 11. </math> Through <math> D, </math> draw a line <math> l_1 </math> [[parallel]] to <math> \overline{AE}, </math> and through <math> E, </math> draw a line <math> l_2 </math> parallel to <math> \overline{AD}. </math> Let <math> F </math> be the | + | [[Equilateral triangle | Equilateral]] <math> \triangle ABC </math> is inscribed in a [[circle]] of [[radius]] <math>2</math>. Extend <math> \overline{AB} </math> through <math> B </math> to point <math> D </math> so that <math> AD=13, </math> and extend <math> \overline{AC} </math> through <math> C </math> to point <math> E </math> so that <math> AE = 11. </math> Through <math> D, </math> draw a line <math> l_1 </math> [[parallel]] to <math> \overline{AE}, </math> and through <math> E, </math> draw a line <math> l_2 </math> parallel to <math> \overline{AD}. </math> Let <math> F </math> be the intersection of <math> l_1 </math> and <math> l_2. </math> Let <math> G </math> be the point on the circle that is [[collinear]] with <math> A </math> and <math> F </math> and distinct from <math> A. </math> Given that the area of <math> \triangle CBG </math> can be expressed in the form <math> \frac{p\sqrt{q}}{r}, </math> where <math> p, q, </math> and <math> r </math> are positive integers, <math> p </math> and <math> r</math> are [[relatively prime]], and <math> q </math> is not [[divisibility | divisible]] by the [[perfect square | square]] of any prime, find <math> p+q+r. </math> |
− | |||
− | == Solution == | + | == Solution 1 == |
− | {{ | + | <center><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></center><!-- Asymptote replacement for Image:Aime2006-2-11.JPG]]--> | ||
+ | Notice that <math>\angle{E} = \angle{BGC} = 120^\circ</math> because <math>\angle{A} = 60^\circ</math>. Also, <math>\angle{GBC} = \angle{GAC} = \angle{FAE}</math> because they both correspond to arc <math>{GC}</math>. So <math>\Delta{GBC} \sim \Delta{EAF}</math>. | ||
+ | <cmath>[EAF] = \frac12 (AE)(EF)\sin \angle AEF = \frac12\cdot11\cdot13\cdot\sin{120^\circ} = \frac {143\sqrt3}4.</cmath> | ||
+ | |||
+ | Because the ratio of the area of two similar figures is the square of the ratio of the corresponding sides, <math>[GBC] = \frac {BC^2}{AF^2}\cdot[EAF] = \frac {12}{11^2 + 13^2 - 2\cdot11\cdot13\cdot\cos120^\circ}\cdot\frac {143\sqrt3}4 = \frac {429\sqrt3}{433}</math>. Therefore, the answer is <math>429+433+3=\boxed{865}</math>. | ||
+ | |||
+ | ==Solution 2: Analytic Geometry/Coord Bash== | ||
+ | |||
+ | Solution by e_power_pi_times_i/edited by srisainandan6 | ||
+ | |||
+ | Let the center of the circle be <math>O</math> and the origin. Then, <math>A (0,2)</math>, <math>B (-\sqrt{3}, -1)</math>, <math>C (\sqrt{3}, -1)</math>. <math>D</math> and <math>E</math> can be calculated easily knowing <math>AD</math> and <math>AE</math>, <math>D (-\dfrac{13}{2}, \dfrac{-13\sqrt{3}+4}{2})</math>, <math>E (\dfrac{11}{2}, \dfrac{-11\sqrt{3}+4}{2})</math>. As <math>DF</math> and <math>EF</math> are parallel to <math>AE</math> and <math>AD</math>, <math>F (-1, -12\sqrt{3}+2)</math>. <math>G</math> and <math>A</math> is the intersection between <math>AF</math> and circle <math>O</math>. Therefore <math>G (-\dfrac{48\sqrt{3}}{433}, -\dfrac{862}{433})</math>. Using the Shoelace Theorem, <math>[CBG] = \dfrac{429\sqrt{3}}{433}</math>, so the answer is <math>\boxed{865}</math>. Note that although the solution may appear short, actually getting all the coordinates take a while as there is plenty of computation. | ||
+ | |||
+ | Note by chrisdiamond10: We can save time calculating the area of the triangle once we have the coordinates of <math>B,C,G</math> by using <math>\frac{b\cdot h}{2}</math>. Use <math>BC</math> as the base, then the base is <math>2\sqrt{3}</math>. The height is easily calculated as <math>-1-\left(-\frac{862}{433}\right)=-1+\frac{862}{433}=\frac{429}{433}</math>, so multiplying base by height and dividing by two we find that the total area is <math>\frac{429\sqrt{3}}{433}</math>, and our answer is <math>\boxed{865}</math>. | ||
+ | |||
+ | == Solution 3: Trig == | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | == Solution 4 == | ||
+ | Note that <math>AB=2\sqrt3</math>, <math>DF=11</math>, and <math>EF=13</math>. If we take a homothety of the parallelogram with respect to <math>A</math>, such that <math>F</math> maps to <math>G</math>, we see that <math>\frac{[ABG]}{[ACG]}=\frac{11}{13}</math>. Since <math>\angle AGB=\angle AGC=60^{\circ}</math>, from the sine area formula we have <math>\frac{BG}{CG}=\frac{11}{13}</math>. Let <math>BG=11k</math> and <math>CG=13k</math>. | ||
+ | |||
+ | By Law of Cosines on <math>\triangle BGC</math>, we have | ||
+ | <cmath>12=k^2(11^2+11\cdot13+13^2)=433k^2\implies k^2=\frac{12}{433}</cmath> | ||
+ | Thus, <math>[CBG]=\frac12 (11k)(13k)\sin 120^{\circ} = \frac{\sqrt3}{4}\cdot 143\cdot \frac{12}{433}=\frac{429\sqrt3}{433}\implies\boxed{865}</math>. | ||
+ | |||
+ | ~rayfish | ||
== See also == | == See also == | ||
{{AIME box|year=2006|n=II|num-b=11|num-a=13}} | {{AIME box|year=2006|n=II|num-b=11|num-a=13}} | ||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 02:36, 19 March 2022
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/Coord 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.
Note by chrisdiamond10: We can save time calculating the area of the triangle once we have the coordinates of by using
. Use
as the base, then the base is
. The height is easily calculated as
, so multiplying base by height and dividing by two we find that the total area is
, and our answer is
.
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 =
.
Solution 4
Note that ,
, and
. If we take a homothety of the parallelogram with respect to
, such that
maps to
, we see that
. Since
, from the sine area formula we have
. Let
and
.
By Law of Cosines on , we have
Thus,
.
~rayfish
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.