2014 AMC 12A Problems/Problem 12

Problem

Two circles intersect at points $A$ and $B$ . The minor arcs $AB$ measure $30^\circ$ on one circle and $60^\circ$ on the other circle. What is the ratio of the area of the larger circle to the area of the smaller circle?

$\textbf{(A) }2\qquad \textbf{(B) }1+\sqrt3\qquad \textbf{(C) }3\qquad \textbf{(D) }2+\sqrt3\qquad \textbf{(E) }4\qquad$

Solution

Let the radius of the larger and smaller circles be $x$ and $y$ , respectively. Also, let their centers be $O_1$ and $O_2$ , respectively. Then the ratio we need to find is $\dfrac{\pi x^2}{\pi y^2} = \dfrac{x^2}{y^2}$ Draw the radii from the centers of the circles to $A$ and $B$ . We can easily conclude that the $30^{\circ}$ belongs to the larger circle, and the $60$ degree arc belongs to the smaller circle. Therefore, $m\angle AO_1B = 30^{\circ}$ and $m\angle AO_2B = 60^{\circ}$ . Note that $\Delta AO_2B$ is equilateral, so when chord AB is drawn, it has length $y$ . Now, applying the Law of Cosines on $\Delta AO_1B$ : $y^2 = x^2 + x^2 - 2x^2\cos{30} = 2x^2 - x^2\sqrt{3} = (2 - \sqrt{3})x^2$ $\dfrac{x^2}{y^2} = \dfrac{1}{2 - \sqrt{3}} = \dfrac{2 + \sqrt{3}}{4-3} = 2 + \sqrt{3} = \boxed{D}$ (Solution by brandbest1)

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2014 AMC 12A Problems/Problem 12

Problem

Solution