Difference between revisions of "2014 AMC 12A Problems/Problem 12"
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+ | ==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 02:33, 26 November 2020
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)
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 |
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