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2014 AMC 12A Problem 12

Problem 12 of 25IntermediateGeometry

Two circles intersect at points AA and B.B. The minor arcs ABAB measure 3030^\circ on one circle and 6060^\circ on the other circle. What is the ratio of the area of the larger circle to the area of the smaller circle?

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Solution

Let the circles have radii RR (with the 3030^\circ arc) and rr (with the 6060^\circ arc). The common chord has length 2Rsin15=2rsin30,2R\sin15^\circ=2r\sin30^\circ, so Rr=sin30sin15.\dfrac{R}{r}=\dfrac{\sin30^\circ}{\sin15^\circ}. The smaller central angle gives the larger radius, so R>r.R\gt r. The area ratio is (Rr)2=14sin215=12(1cos30)=123=2+3. \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.

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Concepts: chord · trigonometry · area ratio

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.