Difference between revisions of "2001 AMC 12 Problems/Problem 18"
(New page: == Problem == A circle centered at <math>A</math> with a radius of 1 and a circle centered at <math>B</math> with a radius of 4 are externally tangent. A third circle is tangent to the fi...) |
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== Solution == | == Solution == | ||
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+ | === Solution 1 === | ||
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Now there are two possibilities: either <math>\frac{4-s}s=-2</math>, or <math>\frac{4-s}s=2</math>. In the first case clearly <math>s<0</math>, hence this is not the correct case. (Note: This case corresponds to the other circle that is tangent to both given circles and the <math>x</math> axis - a large circle whose center is somewhere to the left of <math>A</math>.) The second case solves to <math>s=\frac 43</math>. We then have <math>4r = s^2 = \frac {16}9</math>, hence <math>r = \boxed{\frac 49}</math>. | Now there are two possibilities: either <math>\frac{4-s}s=-2</math>, or <math>\frac{4-s}s=2</math>. In the first case clearly <math>s<0</math>, hence this is not the correct case. (Note: This case corresponds to the other circle that is tangent to both given circles and the <math>x</math> axis - a large circle whose center is somewhere to the left of <math>A</math>.) The second case solves to <math>s=\frac 43</math>. We then have <math>4r = s^2 = \frac {16}9</math>, hence <math>r = \boxed{\frac 49}</math>. | ||
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+ | === Solution 2 === | ||
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+ | The horizontal line is the equivalent of a circle of curvature <math>0</math>, thus we can apply [[Descartes' Circle Formula]]. | ||
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+ | The four circles have curvatures <math>0, 1, \frac 14</math>, and <math>\frac 1r</math>. | ||
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+ | We have <math>2(0^2+1^2+\frac {1}{4^2}+\frac{1}{r^2})=(0+1+\frac 14+\frac 1r)^2</math> | ||
+ | |||
+ | Simplifying, we get <math>\frac{34}{16}+\frac{2}{r^2}=\frac{25}{16}+\frac{5}{2r}+\frac{1}{r^2}</math> | ||
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+ | <math>\frac{1}{r^2}-\frac{5}{2r}+\frac{9}{16}=0</math> | ||
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+ | <math>\frac{16}{r^2}-\frac{40}{r}+9=0</math> | ||
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+ | <math>(\frac{4}{r}-9)(\frac{4}{r}-1)=0</math> | ||
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+ | Obviously <math>r</math> cannot equal <math>4</math>, therefore <math>r = \boxed{\frac 49}</math>. | ||
== See Also == | == See Also == | ||
{{AMC12 box|year=2001|num-b=17|num-a=19}} | {{AMC12 box|year=2001|num-b=17|num-a=19}} |
Revision as of 03:17, 26 January 2012
Problem
A circle centered at with a radius of 1 and a circle centered at with a radius of 4 are externally tangent. A third circle is tangent to the first two and to one of their common external tangents as shown. The radius of the third circle is
Solution
Solution 1
In the triangle we have and , thus by the Pythagorean theorem we have .
We can now pick a coordinate system where the common tangent is the axis and lies on the axis. In this coordinate system we have and .
Let be the radius of the small circle, and let be the -coordinate of its center . We then know that , as the circle is tangent to the axis. Moreover, the small circle is tangent to both other circles, hence we have and .
We have and . Hence we get the following two equations:
Simplifying both, we get
As in our case both and are positive, we can divide the second one by the first one to get .
Now there are two possibilities: either , or . In the first case clearly , hence this is not the correct case. (Note: This case corresponds to the other circle that is tangent to both given circles and the axis - a large circle whose center is somewhere to the left of .) The second case solves to . We then have , hence .
Solution 2
The horizontal line is the equivalent of a circle of curvature , thus we can apply Descartes' Circle Formula.
The four circles have curvatures , and .
We have
Simplifying, we get
Obviously cannot equal , therefore .
See Also
2001 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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 |