Difference between revisions of "2006 AIME II Problems/Problem 12"
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== Solution == | == Solution == | ||
− | {{ | + | 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}</math>~<math>\Delta{EAF}</math>. |
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+ | <math>[EAF] = \frac12\cdot11\cdot13\cdot\sin{120^\circ} = \frac {143\sqrt3}4</math>. | ||
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+ | <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>. | ||
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+ | <math>429+433+3=\boxed{865}</math> | ||
== See also == | == See also == |
Revision as of 09:50, 3 December 2007
Problem
Equilateral is inscribed in a circle of radius 2. 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 ~.
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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 |