Difference between revisions of "2004 AMC 12A Problems/Problem 19"

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label("\(B\)", B,W);
 
label("\(B\)", B,W);
 
label("\(C\)", C,E);
 
label("\(C\)", C,E);
label("\(E\)", E,S);
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label("\(E\)", E,SE);
 
label("\(1\)",(-.4,.7));
 
label("\(1\)",(-.4,.7));
 
label("\(1\)",(0,0.5),W);
 
label("\(1\)",(0,0.5),W);

Revision as of 15:06, 15 July 2012

Problem 19

Circles $A, B$ and $C$ are externally tangent to each other, and internally tangent to circle $D$. Circles $B$ and $C$ are congruent. Circle $A$ has radius $1$ and passes through the center of $D$. What is the radius of circle $B$?

[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]

$\text {(A)} \frac23 \qquad \text {(B)} \frac {\sqrt3}{2} \qquad \text {(C)}\frac78 \qquad \text {(D)}\frac89 \qquad \text {(E)}\frac {1 + \sqrt3}{3}$

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,SE); 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]

Note that $BD= 2-r$ since $D$ is the center of the larger circle of radius $2$. Using the Pythagorean Theorem on $\triangle BDE$,

\begin{align*} r^2 + h^2 &= (2-r)^2 \\ r^2 + h^2 &= 4 - 4r + r^2 \\ h^2 &= 4 - 4r \\ h &= 2\sqrt{1-r} \end{align*}

Now using the Pythagorean Theorem on $\triangle BAE$,

\begin{align*} r^2 + (h+1)^2 &= (r+1)^2 \\ r^2 + h^2 + 2h + 1 &= r^2 + 2r + 1 \\ h^2 + 2h &= 2r \end{align*}

Substituting $h$,

\begin{align*} (4-4r) + 4\sqrt{1-r} &= 2r \\ 4\sqrt{1-r} &= 6r - 4 \\ 16-16r &= 36r^2 - 48r + 16 \\ 0 &= 36r^2 - 32r \\ r &= \frac{32}{36} = \frac{8}{9} \Longrightarrow \qquad \textbf{(D)} \end{align*}

Solution 2

We can apply Descartes' Circle Formula.

The four circles have curvatures $-\frac{1}{2}, 1, \frac{1}{r}$, and $\frac{1}{r}$.

We have $2((-\frac{1}{2})^2+1^2+\frac {1}{r^2}+\frac{1}{r^2})=(-\frac{1}{2}+1+\frac{1}{r}+\frac{1}{r})^2$

Simplifying, we get $\frac{10}{4}+\frac{4}{r^2}=\frac{1}{4}+\frac{2}{r}+\frac{4}{r^2}$

$\frac{2}{r}=\frac{9}{4}$

$r=\frac{8}{9} \Longrightarrow \qquad \textbf{(D)}$

See Also

2004 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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All AMC 12 Problems and Solutions
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