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Difference between revisions of "2014 AMC 12A Problems/Problem 12"

(Created page with "==Problem== Two circles intersect at points <math>A</math> and <math>B</math>. The minor arcs <math>AB</math> measure <math>30^\circ</math> on one circle and <math>60^\circ</ma...")
 
(→‎Solution)
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Draw the radii from the centers of the circles to <math>A</math> and <math>B</math>. We can easily conclude that the <math>30^{\circ}</math> belongs to the larger circle, and the <math>60</math> degree arc belongs to the smaller circle. Therefore, <math>m\angle AO_1B = 30^{\circ}</math> and <math>m\angle AO_2B = 60^{\circ}</math>. Note that <math>\Delta AO_2B</math> is equilateral, so when chord AB is drawn, it has length <math>y</math>. Now, applying the Law of Cosines on <math>\Delta AO_1B</math>:
 
Draw the radii from the centers of the circles to <math>A</math> and <math>B</math>. We can easily conclude that the <math>30^{\circ}</math> belongs to the larger circle, and the <math>60</math> degree arc belongs to the smaller circle. Therefore, <math>m\angle AO_1B = 30^{\circ}</math> and <math>m\angle AO_2B = 60^{\circ}</math>. Note that <math>\Delta AO_2B</math> is equilateral, so when chord AB is drawn, it has length <math>y</math>. Now, applying the Law of Cosines on <math>\Delta AO_1B</math>:
 
<cmath> y^2 = x^2 + x^2 - 2x^2\cos{30} = 2x^2 - x^2\sqrt{3} = (2 - \sqrt{3})x^2 </cmath>
 
<cmath> y^2 = x^2 + x^2 - 2x^2\cos{30} = 2x^2 - x^2\sqrt{3} = (2 - \sqrt{3})x^2 </cmath>
−
<cmath> \dfrac{x^2}{y^2} = \dfrac{1}{2 - \sqrt{3}} = \dfrac{2 + \sqrt{3}}{4-3} = 2 + \sqrt{3} = \boxed{D} </cmath>
+
<cmath> \dfrac{x^2}{y^2} = \dfrac{1}{2 - \sqrt{3}} = \dfrac{2 + \sqrt{3}}{4-3} = 2 + \sqrt{3} = \textbf{(D)} </cmath>
 
(Solution by brandbest1)
 
(Solution by brandbest1)

Revision as of 19:25, 7 February 2014

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

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