Difference between revisions of "2004 AMC 12A Problems/Problem 19"
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==Solution== | ==Solution== | ||
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+ | === Solution 1 === | ||
+ | |||
<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> | ||
+ | |||
+ | === Solution 2 === | ||
+ | |||
+ | We can apply [[Descartes' Circle Formula]]. | ||
+ | |||
+ | The four circles have curvatures <math>-\frac{1}{2}, 1, \frac{1}{r}</math>, and <math>\frac{1}{r}</math>. | ||
+ | |||
+ | 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> | ||
+ | |||
+ | Simplifying, we get <math>\frac{10}{4}+\frac{4}{r^2}=\frac{1}{4}+\frac{2}{r}+\frac{4}{r^2}</math> | ||
+ | |||
+ | <math>\frac{2}{r}=\frac{9}{4}</math> | ||
+ | |||
+ | <math>r=\frac{8}{9} \Longrightarrow \qquad \textbf{(D)}</math> | ||
==See Also== | ==See Also== |
Revision as of 15:15, 29 January 2012
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
?
![[asy] unitsize(15mm); pair A=(-1,0),B=(2/3,8/9),C=(2/3,-8/9),D=(0,0); draw(Circle(D,2)); draw(Circle(A,1)); draw(Circle(B,8/9)); draw(Circle(C,8/9)); label("\(A\)", A); label("\(B\)", B); label("\(C\)", C); label("D", (-1.2,1.8)); [/asy]](http://latex.artofproblemsolving.com/d/b/0/db04d82b5f5ca4618bfc01360dadf7cd5388f624.png)
Solution
Solution 1
![[asy] unitsize(15mm); pair A=(0,1),B=(-8/9,-2/3),C=(8/9,-2/3),D=(0,0), E=(0,-2/3); draw(Circle(D,2)); draw(Circle(A,1)); draw(Circle(B,8/9)); draw(Circle(C,8/9)); draw(A--B--C--A); draw(B--D--C); draw(A--E); dot(A);dot(B);dot(C);dot(D);dot(E); label("\(D\)", D,NW); label("\(A\)", A,N); label("\(B\)", B,W); label("\(C\)", C,E); label("\(E\)", E,S); label("\(1\)",(-.4,.7)); label("\(1\)",(0,0.5),W); label("\(r\)", (-.8,-.1)); label("\(r\)", (-4/9,-2/3),S); label("\(h\)", (0,-1/3), W); [/asy]](http://latex.artofproblemsolving.com/9/d/5/9d510f2c0ec488e38a3dce38ae28d1720d28f06d.png)
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 |
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 |