Difference between revisions of "2014 AMC 12A Problems/Problem 12"
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<cmath> \dfrac{x^2}{y^2} = \dfrac{1}{2 - \sqrt{3}} = \dfrac{2 + \sqrt{3}}{4-3} = 2 + \sqrt{3}=\boxed{\textbf{(D)}}</cmath> | <cmath> \dfrac{x^2}{y^2} = \dfrac{1}{2 - \sqrt{3}} = \dfrac{2 + \sqrt{3}}{4-3} = 2 + \sqrt{3}=\boxed{\textbf{(D)}}</cmath> | ||
(Solution by brandbest1) | (Solution by brandbest1) | ||
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+ | ==See Also== | ||
+ | {{AMC12 box|year=2014|ab=A|num-b=11|num-a=13}} | ||
+ | {{MAA Notice}} |
Revision as of 13:31, 8 February 2014
Problem
Two circles intersect at points and . The minor arcs measure on one circle and on the other circle. What is the ratio of the area of the larger circle to the area of the smaller circle?
Solution
Let the radius of the larger and smaller circles be and , respectively. Also, let their centers be and , respectively. Then the ratio we need to find is Draw the radii from the centers of the circles to and . We can easily conclude that the belongs to the larger circle, and the degree arc belongs to the smaller circle. Therefore, and . Note that is equilateral, so when chord AB is drawn, it has length . Now, applying the Law of Cosines on : (Solution by brandbest1)
See Also
2014 AMC 12A (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 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.