Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2006 AIME II Problems/Problem 12"

 
 
(20 intermediate revisions by 12 users not shown)
Line 1: Line 1:
−
#REDIRECT [[2006 AIME A Problems/Problem 12]]
+
== Problem ==
 +
[[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 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 ==
 +
{{AIME box|year=2006|n=II|num-b=11|num-a=13}}
 +
 
 +
[[Category:Intermediate Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 01:36, 19 March 2022

Problem

Equilateral $\triangle ABC$ is inscribed in a circle of radius $2$. Extend $\overline{AB}$ through $B$ to point $D$ so that $AD=13,$ and extend $\overline{AC}$ through $C$ to point $E$ so that $AE = 11.$ Through $D,$ draw a line $l_1$ parallel to $\overline{AE},$ and through $E,$ draw a line $l_2$ parallel to $\overline{AD}.$ Let $F$ be the intersection of $l_1$ and $l_2.$ Let $G$ be the point on the circle that is collinear with $A$ and $F$ and distinct from $A.$ Given that the area of $\triangle CBG$ can be expressed in the form $\frac{p\sqrt{q}}{r},$ where $p, q,$ and $r$ are positive integers, $p$ and $r$ are relatively prime, and $q$ is not divisible by the square of any prime, find $p+q+r.$

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]

Notice that $\angle{E} = \angle{BGC} = 120^\circ$ because $\angle{A} = 60^\circ$. Also, $\angle{GBC} = \angle{GAC} = \angle{FAE}$ because they both correspond to arc ${GC}$. So $\Delta{GBC} \sim \Delta{EAF}$.

\[[EAF] = \frac12 (AE)(EF)\sin \angle AEF = \frac12\cdot11\cdot13\cdot\sin{120^\circ} = \frac {143\sqrt3}4.\]

Because the ratio of the area of two similar figures is the square of the ratio of the corresponding sides, $[GBC] = \frac {BC^2}{AF^2}\cdot[EAF] = \frac {12}{11^2 + 13^2 - 2\cdot11\cdot13\cdot\cos120^\circ}\cdot\frac {143\sqrt3}4 = \frac {429\sqrt3}{433}$. Therefore, the answer is $429+433+3=\boxed{865}$.

Solution 2: Analytic Geometry/Coord Bash

Solution by e_power_pi_times_i/edited by srisainandan6

Let the center of the circle be $O$ and the origin. Then, $A (0,2)$, $B (-\sqrt{3}, -1)$, $C (\sqrt{3}, -1)$. $D$ and $E$ can be calculated easily knowing $AD$ and $AE$, $D (-\dfrac{13}{2}, \dfrac{-13\sqrt{3}+4}{2})$, $E (\dfrac{11}{2}, \dfrac{-11\sqrt{3}+4}{2})$. As $DF$ and $EF$ are parallel to $AE$ and $AD$, $F (-1, -12\sqrt{3}+2)$. $G$ and $A$ is the intersection between $AF$ and circle $O$. Therefore $G (-\dfrac{48\sqrt{3}}{433}, -\dfrac{862}{433})$. Using the Shoelace Theorem, $[CBG] = \dfrac{429\sqrt{3}}{433}$, so the answer is $\boxed{865}$. 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 $B,C,G$ by using $\frac{b\cdot h}{2}$. Use $BC$ as the base, then the base is $2\sqrt{3}$. The height is easily calculated as $-1-\left(-\frac{862}{433}\right)=-1+\frac{862}{433}=\frac{429}{433}$, so multiplying base by height and dividing by two we find that the total area is $\frac{429\sqrt{3}}{433}$, and our answer is $\boxed{865}$.

Solution 3: Trig

Lines $l_1$ and $l_2$ are constructed such that $AEFD$ is a parallelogram, hence $DF = 13$. Since $BAC$ is equilateral with angle of $60^{\circ}$, angle $D$ is $120^{\circ}$. Use law of cosines to find $AF = \sqrt{433}$. Then use law of sines to find angle $BAG$ and $GAC$. Next we use Ptolemy's Theorem on $ABGC$ to find that $CG + BG = AG$. Next we use law of cosine on triangles $BAG$ and $GAC$, solving for BG and CG respectively. Subtract the two equations and divide out a $BG + CG$ to find the value of $CG - BG$. Next, $AG = 2\cdot R \cos{\theta}$, where R is radius of circle $= 2$ and $\theta =$ angle $BAG$. We already know sine of the angle so find cosine, hence we have found $AG$. At this point it is system of equation yielding $CG = \frac{26\sqrt{3}}{\sqrt{433}}$ and $BG = \frac{22\sqrt{3}}{\sqrt{433}}$. Given $[CBG] = \frac{BC \cdot CG \cdot BG}{4R}$, and $BC = 2\sqrt{3}$ by $30-60-90$ triangle, we can evaluate to find $[CBG] = \frac{429\sqrt{3}}{433}$, to give answer = $\boxed{865}$.

Solution 4

Note that $AB=2\sqrt3$, $DF=11$, and $EF=13$. If we take a homothety of the parallelogram with respect to $A$, such that $F$ maps to $G$, we see that $\frac{[ABG]}{[ACG]}=\frac{11}{13}$. Since $\angle AGB=\angle AGC=60^{\circ}$, from the sine area formula we have $\frac{BG}{CG}=\frac{11}{13}$. Let $BG=11k$ and $CG=13k$.

By Law of Cosines on $\triangle BGC$, we have \[12=k^2(11^2+11\cdot13+13^2)=433k^2\implies k^2=\frac{12}{433}\] Thus, $[CBG]=\frac12 (11k)(13k)\sin 120^{\circ} = \frac{\sqrt3}{4}\cdot 143\cdot \frac{12}{433}=\frac{429\sqrt3}{433}\implies\boxed{865}$.

~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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems/Problem_12&oldid=172753"