2009 AIME II Problems/Problem 5
Contents
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 angle degrees because we are given that triangle is equilateral. Using the Law of Cosines on triangle , we obtain
.
The and the terms cancel out:
. The radius of circle is , so the answer is .
Video Solution
https://www.youtube.com/watch?v=KKVxQV4hszo&t=7s
See Also
2009 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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