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

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\textbf{(D) }2+\sqrt3\qquad
 
\textbf{(D) }2+\sqrt3\qquad
 
\textbf{(E) }4\qquad</math>
 
\textbf{(E) }4\qquad</math>
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[[Category: Introductory Geometry Problems]]
  
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==Solution==
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==Solution 1==
  
 
Let the radius of the larger and smaller circles be <math>x</math> and <math>y</math>, respectively. Also, let their centers be <math>O_1</math> and <math>O_2</math>, respectively. Then the ratio we need to find is
 
Let the radius of the larger and smaller circles be <math>x</math> and <math>y</math>, respectively. Also, let their centers be <math>O_1</math> and <math>O_2</math>, respectively. Then the ratio we need to find is
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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>
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<cmath> \dfrac{x^2}{y^2} = \dfrac{1}{2 - \sqrt{3}} = \dfrac{2 + \sqrt{3}}{4-3} = 2 + \sqrt{3} = \textbf{(D)} </cmath>
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<cmath> \dfrac{x^2}{y^2} = \dfrac{1}{2 - \sqrt{3}} = \dfrac{2 + \sqrt{3}}{4-3} = 2 + \sqrt{3}=\boxed{\textbf{(D)}}</cmath>
 
(Solution by brandbest1)
 
(Solution by brandbest1)
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==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{\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>
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(Solution by kevin38017)
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==Solution 3==
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Let the radius of the smaller and larger circle be <math>r</math> and <math>R</math>, respectively. We see that half the length of the chord is equal to <math>r \sin 30^{\circ}</math>, which is also equal to <math>R \sin 15^{\circ}</math>. Recall that <math>\sin 15^{\circ} = \frac{\sqrt{6} - \sqrt{2}}{4}</math> and <math>\sin 30^{\circ} = \frac{1}{2}</math>. From this, we get <math>r = \frac{\sqrt{6} - \sqrt{2}}{2} R</math>, or <math>r^2 = \frac{8 - 2 \sqrt{12}}{4} R^2 = \left(2 - \sqrt{3}\right) R^2</math>, which is equivalent to <math>R^2 = \left(2 + \sqrt{3}\right) r^2</math>.
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(Solution by soy_un_chemisto)
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==Solution 4==
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As in the previous solutions let the radius of the smaller and larger circles be <math>r</math> and <math>R</math>, respectively. Also, let their centers be <math>O_1</math> and <math>O_2</math>, respectively. Now draw two congruent chords from points <math>A</math> and <math>B</math> to the end of the smaller circle, creating an isosceles triangle. Label that point <math>X</math>. Recalling the Inscribed Angle Theorem, we then see that <math>m\angle AXB = \frac{m\angle AO_1B}{2} = 30^{\circ}= m\angle AO_2B</math>. Based on this information, we can conclude that triangles <math>AXB</math> and  <math>AO_2B</math> are congruent via ASA Congruence.
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Next draw the height of <math>AXB</math> from <math>X</math> to <math>AB</math>. Note we've just created a right triangle with hypotenuse <math>R</math>, base <math>\frac{r}{2}</math>, and height <math>\frac{r\sqrt{3}}{2} + r</math>
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Thus using the Pythagorean Theorem we can express <math>R^2</math> in terms of <math>r</math>
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<cmath>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})</cmath>
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We can now determine the ratio between the larger and smaller circles:
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<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>
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(Solution by derekxu)
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==Solution 5==
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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:
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<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>
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Then we can find the ratio:
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<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>
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(Solution by weirdo)
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==Solution 6 (using answer choices)==
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We will estimate the answer using a wrong method then guess the correct answer choice.
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Let the radius of the larger and smaller circles be <math>R</math> and <math>r</math>, respectively. Pretend line <math>AB</math> is equal to the arc length of both circles, then <cmath>\frac{30^{\circ}}{360^{\circ}}\cdot 2\pi R=\frac{60^{\circ}}{360^{\circ}}\cdot 2\pi r</cmath><cmath> R=2r</cmath> and the answer to the problem is <cmath>\frac{\pi R^2}{\pi r^2} = 4</cmath>
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But, this is only an estimate, the correct answer is not 4, but instead ''about'' 4. We can asume that the closest other answer choice to 4 is correct: <cmath>\boxed{\textbf{(D) }2+\sqrt3}</cmath>
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(Solution by FlareVa)
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==Video Solution==
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https://youtu.be/tdhHKQIWdXY
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==See Also==
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{{AMC12 box|year=2014|ab=A|num-b=11|num-a=13}}
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{{MAA Notice}}

Latest revision as of 22:37, 23 June 2022

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)

Solution 6 (using answer choices)

We will estimate the answer using a wrong method then guess the correct answer choice.

Let the radius of the larger and smaller circles be $R$ and $r$, respectively. Pretend line $AB$ is equal to the arc length of both circles, then \[\frac{30^{\circ}}{360^{\circ}}\cdot 2\pi R=\frac{60^{\circ}}{360^{\circ}}\cdot 2\pi r\]\[R=2r\] and the answer to the problem is \[\frac{\pi R^2}{\pi r^2} = 4\] But, this is only an estimate, the correct answer is not 4, but instead about 4. We can asume that the closest other answer choice to 4 is correct: \[\boxed{\textbf{(D) }2+\sqrt3}\]

(Solution by FlareVa)

Video Solution

https://youtu.be/tdhHKQIWdXY

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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