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

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

Revision as of 23:00, 4 February 2019

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: \[\left(\dfrac {x}{y} \right)^2 = \left(\frac{1}{2}{\dfrac{\sqrt{6} - \sqrt{2}}{4}} \right)^2 = \left(\dfrac{\sqrt{6} + \sqrt{2}}{2} \right)^2 = 2 + \sqrt{3} =\boxed{\textbf{(D)}}\] (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)

See Also

2014 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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