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> | ||
+ | [[Category: Introductory Geometry Problems]] | ||
==Solution 1== | ==Solution 1== | ||
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==Solution 2== | ==Solution 2== | ||
− | 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>. | + | 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> |
(Solution by kevin38017) | (Solution by kevin38017) | ||
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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>. | 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>. | ||
− | ( | + | (Solution by soy_un_chemisto) |
+ | |||
+ | ==Solution 4== | ||
+ | |||
+ | 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. | ||
+ | |||
+ | |||
+ | 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> | ||
+ | Thus using the Pythagorean Theorem we can express <math>R^2</math> in terms of <math>r</math> | ||
+ | <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> | ||
+ | |||
+ | 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> | ||
+ | |||
+ | (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) | ||
+ | |||
+ | ==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 <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> | ||
+ | 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> | ||
+ | |||
+ | (Solution by FlareVa) | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/tdhHKQIWdXY | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2014|ab=A|num-b=11|num-a=13}} | {{AMC12 box|year=2014|ab=A|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 23:37, 23 June 2022
Contents
Problem
Two circles intersect at points and
. The minor arcs
measure
on one circle and
on the other circle. What is the ratio of the area of the larger circle to the area of the smaller circle?
Solution 1
Let the radius of the larger and smaller circles be and
, respectively. Also, let their centers be
and
, respectively. Then the ratio we need to find is
Draw the radii from the centers of the circles to
and
. We can easily conclude that the
belongs to the larger circle, and the
degree arc belongs to the smaller circle. Therefore,
and
. Note that
is equilateral, so when chord AB is drawn, it has length
. Now, applying the Law of Cosines on
:
(Solution by brandbest1)
Solution 2
Again, let the radius of the larger and smaller circles be and
, respectively, and let the centers of these circles be
and
, respectively. Let
bisect segment
. Note that
and
are right triangles, with
and
. We have
and
and
. Since the ratio of the area of the larger circle to that of the smaller circle is simply
, we just need to find
and
. We know
, and we can use the angle sum formula or half angle formula to compute
. Plugging this into the previous expression, we get:
(Solution by kevin38017)
Solution 3
Let the radius of the smaller and larger circle be and
, respectively. We see that half the length of the chord is equal to
, which is also equal to
. Recall that
and
. From this, we get
, or
, which is equivalent to
.
(Solution by soy_un_chemisto)
Solution 4
As in the previous solutions let the radius of the smaller and larger circles be and
, respectively. Also, let their centers be
and
, respectively. Now draw two congruent chords from points
and
to the end of the smaller circle, creating an isosceles triangle. Label that point
. Recalling the Inscribed Angle Theorem, we then see that
. Based on this information, we can conclude that triangles
and
are congruent via ASA Congruence.
Next draw the height of from
to
. Note we've just created a right triangle with hypotenuse
, base
, and height
Thus using the Pythagorean Theorem we can express
in terms of
We can now determine the ratio between the larger and smaller circles:
(Solution by derekxu)
Solution 5
Let the radius of the larger and smaller circles be and
, respectively, and let the centers of these circles be
and
, respectively. Draw the radii and
, and note that
because
is equilateral. Also,
. Then, mark point
inside the larger circle such that
is a right angle and
. Notice that
is a 45-45-90 triangle, so
. Now extend
to
, and label the intersection
. Since
,
so this creates 30-60-90 triangle
. Therefore,
and
. We can also see that
, so
is an isosceles triangle with
. So
. This means:
Then we can find the ratio:
(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 and
, respectively. Pretend line
is equal to the arc length of both circles, then
and the answer to the problem is
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:
(Solution by FlareVa)
Video Solution
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.