2009 AIME II Problems/Problem 5
Problem 5
Equilateral triangle is inscribed in circle , which has radius . Circle with radius is internally tangent to circle at one vertex of . Circles and , both with radius , are internally tangent to circle at the other two vertices of . Circles , , and are all externally tangent to circle , which has radius , where and are relatively prime positive integers. Find .
Solution
Let be the intersection of the circles with centers and , and be the intersection of the circles with centers and . Since the radius of is , = . Assume = . Then and are radii of circle and have length . = , and it can easily be shown that angle = degrees. Using the Law of Cosines on triangle , we obtain
= + - cos .
The and the cos cancel out:
+ + = + -
+ = -
= = . The radius of circle is + = , so the answer is + = .
See Also
2009 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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