# 2004 AMC 12A Problems/Problem 19

The following problem is from both the 2004 AMC 12A #19 and 2004 AMC 10A #23, so both problems redirect to this page.

## 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 $[asy] import graph; size(400); defaultpen(fontsize(10)); pair OA=(-1,0),OB=(2/3,8/9),OC=(2/3,-8/9),OD=(0,0),E=(2/3,0); real t = 2.5; pair OA1=(-2+2*t,0),OB1=(4/3+2*t,16/9),OC1=(4/3+2*t,-16/9),OD1=(0+2*t,0),E1=(4/3+2*t,0); draw(Circle(OD,2)); draw(Circle(OA,1)); draw(Circle(OB,8/9)); draw(Circle(OC,8/9)); draw(OA--OB--OC--cycle); draw(OD--OB--OB+(OB-OD)*4/5); draw(OA--E); label("O_{A}",OA,(-1,0)); label("O_{B}",OB,(-1,1)); label("O_{C}",OC,(-1,-1)); label("O_{D}",OD,(-1,-1)); label("E",E,(0.5,-1)); label("r",OB+(OB-OD)*2/5,(-0.5,1)); label("r",(1*OA+3*OB)/4,(-0.5,1)); dot(OA^^OB^^OC^^OD^^E); draw(OA1--OB1--OC1--cycle); draw(OD1--OB1); draw(OA1--E1); label("O_{A}",OA1,(-1,0)); label("O_{B}",OB1,(1,1)); label("O_{C}",OC1,(1,-1)); label("O_{D}",OD1,(0,-1)); label("E",E1,(1,0)); label("1+r",(OA1+OB1)/2,(-0.5,1)); label("r",(E1+OB1)/2,(1,0)); label("r",(E1+OC1)/2,(1,0)); label("2-r",(OB1+OD1)/2,(-1,0)); label("1",(OA1+OD1)/2,(0,-1)); label("x",(E1+OD1)/2,(0,-1)); dot(OA1^^OB1^^OC1^^OD1^^E1); [/asy]$ Let $O_{i}$ be the center of circle $i$ for all $i \in \{A,B,C,D\}$ and let $E$ be the tangent point of $B,C$. Since the radius of $D$ is the diameter of $A$, the radius of $D$ is $2$. Let the radius of $B,C$ be $r$ and let $O_{D}E = x$. If we connect $O_{A},O_{B},O_{C}$, we get an isosceles triangle with lengths $1 + r, 2r$. Then right triangle $O_{D}O_{B}O_{E}$ has legs $r, x$ and hypotenuse $2-r$. Solving for $x$, we get $x^2 = (2-r)^2 - r^2 \Longrightarrow x = \sqrt{4-4r}$.

Also, right triangle $O_{A}O_{B}O_{E}$ has legs $r, 1+x$, and hypotenuse $1+r$. Solving, $\begin{eqnarray*} r^2 + (1+\sqrt{4-4r})^2 &=& (1+r)^2\\ 1+4-4r+2\sqrt{4-4r}&=& 2r + 1\\ 1-r &=& \left(\frac{6r-4}{4}\right)^2\\ \frac{9}{4}r^2-2r&=& 0\\ r &=& \frac 89 \end{eqnarray*}$

So the answer is $\boxed{\mathrm{(D)}\ \frac{8}{9}}$.

### Solution 2 $[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 3

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 $\frac{10}{4}+\frac{4}{r^2}=\frac{1}{4}+\frac{2}{r}+\frac{4}{r^2}$ $\frac{2}{r}=\frac{9}{4}$ $r=\frac{8}{9} \Longrightarrow \qquad \textbf{(D)}$

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