Difference between revisions of "2004 AMC 12A Problems/Problem 19"
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==Solution== | ==Solution== | ||
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+ | === Solution 1 === | ||
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<center><asy> | <center><asy> | ||
unitsize(15mm); | unitsize(15mm); | ||
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0 &= 36r^2 - 32r \ | 0 &= 36r^2 - 32r \ | ||
r &= \frac{32}{36} = \frac{8}{9} \Longrightarrow \qquad \textbf{(D)} \end{align*}</cmath> | r &= \frac{32}{36} = \frac{8}{9} \Longrightarrow \qquad \textbf{(D)} \end{align*}</cmath> | ||
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+ | === Solution 2 === | ||
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+ | We can apply [[Descartes' Circle Formula]]. | ||
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+ | The four circles have curvatures <math>-\frac{1}{2}, 1, \frac{1}{r}</math>, and <math>\frac{1}{r}</math>. | ||
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+ | We have <math>2((-\frac{1}{2})^2+1^2+\frac {1}{r^2}+\frac{1}{r^2})=(-\frac{1}{2}+1+\frac{1}{r}+\frac{1}{r})^2</math> | ||
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+ | Simplifying, we get <math>\frac{10}{4}+\frac{4}{r^2}=\frac{1}{4}+\frac{2}{r}+\frac{4}{r^2}</math> | ||
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+ | <math>\frac{2}{r}=\frac{9}{4}</math> | ||
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+ | <math>r=\frac{8}{9} \Longrightarrow \qquad \textbf{(D)}</math> | ||
==See Also== | ==See Also== |
Revision as of 14:15, 29 January 2012
Contents
[hide]Problem 19
Circles and are externally tangent to each other, and internally tangent to circle . Circles and are congruent. Circle has radius and passes through the center of . What is the radius of circle ?
Solution
Solution 1
Note that since is the center of the larger circle of radius . Using the Pythagorean Theorem on ,
Now using the Pythagorean Theorem on ,
Substituting ,
Solution 2
We can apply Descartes' Circle Formula.
The four circles have curvatures , and .
We have
Simplifying, we get
See Also
2004 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
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All AMC 12 Problems and Solutions |