2009 AMC 10B Problems/Problem 16
Problem
Points and lie on a circle centered at , each of and are tangent to the circle, and is equilateral. The circle intersects at . What is ?
Solution
Solution 1
As is equilateral, we have , hence . Then , and from symmetry we have . Thus, this gives us .
We know that , as lies on the circle. From we also have , Hence , therefore , and .
Solution 2
As in the previous solution, we find out that . Hence and are both equilateral.
We then have , hence is the incenter of , and as is equilateral, is also its centroid. Hence , and as , we have , therefore , and as before we conclude that .
Solution 3
by SSS congruence, so . Since is tangent to the circle, it is perpendicular to . This means that is a 30-60-90 triangle. The ratio of the side-lengths of a 30-60-90 triangle is , so . . Since is the radius of the circle, and . Hence, , and the answer is ~azc1027
See Also
2009 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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