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

Revision as of 19:18, 9 October 2021

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

Solution 5

Let the radius of the larger and smaller circles be $R$ and $r$ , respectively, and let the centers of these circles be $O_1$ and $O_2$ , respectively. Draw the radii and $AB$ , and note that $AB=r$ because $\triangle ABO_1$ is equilateral. Also, $m\angle O_1AB = m\angle O_1BA = \frac{180^{\circ}-30^{\circ}}{2} = 75^{\circ}$ . Then, mark point $X$ inside the larger circle such that $\angle AXB$ is a right angle and $AX = BX$ . Notice that $\triangle ABX$ is a 45-45-90 triangle, so $AX = BX = \frac{r\sqrt{2}}{2}$ . Now extend $BX$ to $AO_1$ , and label the intersection $Y$ . Since $m\angle YAB = 75^{\circ}$ , $m\angle YAX = 75^{\circ} - 45^{\circ} = 30^{\circ}$ so this creates 30-60-90 triangle $\triangle AYX$ . Therefore, $YX = \frac{r\sqrt{6}}{6}$ and $AY = \frac{r\sqrt{6}}{3}$ . We can also see that $\angle YBO_1 = 75^{\circ}-45^{\circ} = 30^{\circ}=\angle YO_1B$ , so $\triangle YO_1B$ is an isosceles triangle with $YB = YO_1$ . So $YO_1 = \frac{r\sqrt{6}}{6} + \frac{r\sqrt{2}}{2}$ . This means: $R = AO_1 = \frac{r\sqrt{6}}{6} + \frac{r\sqrt{2}}{2} + \frac{r\sqrt{6}}{3} = \frac{r(\sqrt{6} + \sqrt{2})}{2}$ Then we can find the ratio: $\frac{\pi R^2}{\pi r^2} = \frac{\pi [\frac{r(\sqrt{6} + \sqrt{2})}{2}]^2}{\pi r^2} = \frac{(\sqrt{6}+\sqrt{2})}{4}=\frac{8+4\sqrt{3}}{4}=\boxed{\textbf{(D) } 2 + \sqrt{3}}$

(Solution by weirdo)

Video Solution

https://youtu.be/tdhHKQIWdXY

@@ Line 40: / Line 40: @@
 We can now determine the ratio between the larger and smaller circles:
-<cmath>\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}}</cmath>
+<cmath>\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}}</cmath>
 (Solution by derekxu)
+==Solution 5==
+Let the radius of the larger and smaller circles be <math>R</math> and <math>r</math>, respectively, and let the centers of these circles be <math>O_1</math> and <math>O_2</math>, respectively. Draw the radii and <math>AB</math>, and note that <math>AB=r</math> because <math>\triangle ABO_1</math> is equilateral. Also, <math>m\angle O_1AB = m\angle O_1BA = \frac{180^{\circ}-30^{\circ}}{2} = 75^{\circ}</math>. Then, mark point <math>X</math> inside the larger circle such that <math>\angle AXB</math> is a right angle and <math>AX = BX</math>. Notice that <math>\triangle ABX</math> is a 45-45-90 triangle, so <math>AX = BX = \frac{r\sqrt{2}}{2}</math>. Now extend <math>BX</math> to <math>AO_1</math>, and label the intersection <math>Y</math>. Since <math>m\angle YAB = 75^{\circ}</math>, <math>m\angle YAX = 75^{\circ} - 45^{\circ} = 30^{\circ}</math> so this creates 30-60-90 triangle <math>\triangle AYX</math>. Therefore, <math>YX = \frac{r\sqrt{6}}{6}</math> and <math>AY = \frac{r\sqrt{6}}{3}</math>. We can also see that <math>\angle YBO_1 = 75^{\circ}-45^{\circ} = 30^{\circ}=\angle YO_1B</math>, so <math>\triangle YO_1B</math> is an isosceles triangle with <math>YB = YO_1</math>. So <math>YO_1 = \frac{r\sqrt{6}}{6} + \frac{r\sqrt{2}}{2}</math>. This means:
+<cmath>R = AO_1 = \frac{r\sqrt{6}}{6} + \frac{r\sqrt{2}}{2} + \frac{r\sqrt{6}}{3} = \frac{r(\sqrt{6} + \sqrt{2})}{2}</cmath>
+Then we can find the ratio:
+<cmath>\frac{\pi R^2}{\pi r^2} = \frac{\pi [\frac{r(\sqrt{6} + \sqrt{2})}{2}]^2}{\pi r^2} = \frac{(\sqrt{6}+\sqrt{2})}{4}=\frac{8+4\sqrt{3}}{4}=\boxed{\textbf{(D) }  2 + \sqrt{3}}</cmath>
+(Solution by weirdo)
 ==Video Solution==

2014 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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