2014 AMC 12A Problem 12

Question

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?

2

1+\sqrt3

3

2+\sqrt3

4

The common chord

AB

equals

2R\sin15^\circ

in one circle and

2r\sin30^\circ

in the other

The area ratio is

\left(\dfrac{R}{r}\right)^2=\dfrac{1}{4\sin^2 15^\circ}

, and

4\sin^2 15^\circ=2-\sqrt3

Let the circles have radii

R

(with the

30^\circ

arc) and

r

(with the

60^\circ

arc). The common chord has length

2R\sin15^\circ=2r\sin30^\circ,

so

\dfrac{R}{r}=\dfrac{\sin30^\circ}{\sin15^\circ}.

The smaller central angle gives the larger radius, so

R\gt r.

The area ratio is

\begin{gathered} \left(\dfrac{R}{r}\right)^2\\ =\dfrac{1}{4\sin^2 15^\circ}\\ =\dfrac{1}{2(1-\cos30^\circ)}\\ =\dfrac{1}{2-\sqrt3}=2+\sqrt3. \end{gathered}

Thus, the correct answer is D.

Problem 12 in other years