Difference between revisions of "2006 AIME II Problems/Problem 12"
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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>. | ||
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+ | ==Solution 2: Analytic Geometry== | ||
+ | |||
+ | Solution by e_power_pi_times_i | ||
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+ | 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> | ||
== See also == | == See also == |
Revision as of 17:05, 28 July 2016
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
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
Solution by e_power_pi_times_i
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
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.