# 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 1

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 $\Delta AO_1B$: $$y^2 = x^2 + x^2 - 2x^2\cos{30} = 2x^2 - x^2\sqrt{3} = (2 - \sqrt{3})x^2$$ $$\dfrac{x^2}{y^2} = \dfrac{1}{2 - \sqrt{3}} = \dfrac{2 + \sqrt{3}}{4-3} = 2 + \sqrt{3}=\boxed{\textbf{(D)}}$$ (Solution by brandbest1)

## Solution 2

Again, let the radius of the larger and smaller circles be $x$ and $y$, respectively, and let the centers of these circles be $O_1$ and $O_2$, respectively. Let $X$ bisect segment $AB$. Note that $\triangle AXO_1$ and $\triangle AXO_2$ are right triangles, with $\angle AO_1X=15^{\circ}$ and $\angle AO_2X=30^{\circ}$. We have $\sin{15} = \dfrac{AX}{x}$ and $\sin{30} = \dfrac{AX}{y}$ and $\dfrac{x}{y} = \dfrac{\sin{30}}{\sin{15}}$. Since the ratio of the area of the larger circle to that of the smaller circle is simply $\dfrac{\pi x^2}{\pi y^2} = \left(\dfrac{x}{y} \right)^2 = \left(\dfrac{\sin{30}}{\sin{15}} \right)^2$, we just need to find $\sin{30}$ and $\sin{15}$. We know $\sin{30} = \dfrac{1}{2}$, and we can use the angle sum formula or half angle formula to compute $\sin{15} = \dfrac{\sqrt{6} - \sqrt{2}}{4}$. Plugging this into the previous expression, we get: $$\left(\dfrac {x}{y} \right)^2 = \left(\dfrac{\dfrac{1}{2}}{\dfrac{\sqrt{6} - \sqrt{2}}{4}} \right)^2 = \left(\dfrac{\sqrt{6} + \sqrt{2}}{2} \right)^2 = 2 + \sqrt{3} =\boxed{\textbf{(D)}}$$ (Solution by kevin38017)

## Solution 3

Let the radius of the smaller and larger circle be $r$ and $R$, respectively. We see that half the length of the chord is equal to $r \sin 30^{\circ}$, which is also equal to $R \sin 15^{\circ}$. Recall that $\sin 15^{\circ} = \frac{\sqrt{6} - \sqrt{2}}{4}$ and $\sin 30^{\circ} = \frac{1}{2}$. From this, we get $r = \frac{\sqrt{6} - \sqrt{2}}{2} R$, or $r^2 = \frac{8 - 2 \sqrt{12}}{4} R^2 = \left(2 - \sqrt{3}\right) R^2$, which is equivalent to $R^2 = \left(2 + \sqrt{3}\right) r^2$.

(Solution by soy_un_chemisto)

## Solution 4

As in the previous solutions let the radius of the smaller and larger circles be $r$ and $R$, respectively. Also, let their centers be $O_1$ and $O_2$, respectively. Now draw two congruent chords from points $A$ and $B$ to the end of the smaller circle, creating an isosceles triangle. Label that point $X$. Recalling the Inscribed Angle Theorem, we then see that $m\angle AXB = \frac{m\angle AO_1B}{2} = 30^{\circ}= m\angle AO_2B$. Based on this information, we can conclude that triangles $AXB$ and $AO_2B$ are congruent via ASA Congruence.

Next draw the height of $AXB$ from $X$ to $AB$. Note we've just created a right triangle with hypotenuse $R$, base $\frac{r}{2}$, and height $\frac{r\sqrt{3}}{2} + r$ Thus using the Pythagorean Theorem we can express $R^2$ in terms of $r$ $$R^2 = (\frac{r}{2})^2 + (\frac{r\sqrt{3}}{2} + r)^2 = r^2 + \frac{r^2}{4} + \frac{3r^2}{4} + (2)(\frac{r\sqrt{3}}{2})(r) = 2r^2 + r^2\sqrt{3} = r^2(2 + \sqrt{3})$$

We can now determine the ratio between the larger and smaller circles: $$\frac{Area [O_2]}{Area [O_1]} = \frac{\pi R^2}{\pi r^2} = \frac{\pi r^2(2 + \sqrt{3})}{\pi r^2} =\boxed{\textbf{(D)} 2 + \sqrt{3}}$$

(Solution by derekxu)

## Video Solution

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 