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

2014 AMC 12A Problem 12 — Solution

2014 AMC 12A Problem 12

Below is the professionally curated solution for Problem 12 of the 2014 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2014 AMC 12A solutions, or check the answer key.

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Concepts:chord trigonometry area ratio

Difficulty rating: 1630

12.

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?

$2$

$1+\sqrt3$

$3$

$2+\sqrt3$

$4$

Solution:

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

Thus, the correct answer is D.

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