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

Revision as of 14:15, 29 January 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,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]$

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 $\frac{10}{4}+\frac{4}{r^2}=\frac{1}{4}+\frac{2}{r}+\frac{4}{r^2}$

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

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

@@ Line 20: / Line 20: @@
 ==Solution==
+=== Solution 1 ===
 <center><asy>
 unitsize(15mm);
@@ Line 72: / Line 75: @@
 &= 36r^2 - 32r \\
 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==

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

Page

Toolbox

Search