Difference between revisions of "2006 AIME II Problems/Problem 12"
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[[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> | [[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 == |
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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>. | 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/ | + | ==Solution 2: Analytic Geometry/Coord Bash== |
Solution by e_power_pi_times_i/edited by srisainandan6 | 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 | + | 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 == | == Solution 3: Trig == | ||
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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>. | ||
+ | == 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 | + | [[Category:Intermediate Geometry Problems]] |
{{MAA Notice}} | {{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
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 |
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