2004 AMC 12A Problems/Problem 19

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 $B$ ?

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

Solution

$[asy] unitsize(20mm); pair A=(0,1),B=(-8/9,-2/3),C=(8/9,-2/3),D=(0,0), E=(0,-2/3); draw(Circle(D,2)); draw(Circle(A,1)); draw(Circle(B,8/9)); draw(Circle(C,8/9)); draw(A--B--C--A); draw(B--D--C); draw(A--E); dot(A);dot(B);dot(C);dot(D);dot(E); label("$D$", D,N); label("$A$", A,N); label("$B$", B,W); label("$C$", C,E); label("$E$", E,S); label("$1$",(-.4,.7)); label("$1$",(0,.5),E); label("$r$", (-.8,-.1)); label("$r$", (-4/9,-2/3),S); label("$h$", (0,-1/3),E); [/asy]$

Note that $BD= 2-r$ since D is the center of the larger circle of radius 2

Using the Pythagorean Theorem on $\triangle BDE$

$r^2 + h^2 = (2-r)^2$

$r^2 + h^2 = 4 - 4r + r^2$

$h^2 = 4 - 4r$

$h = 2\sqrt{1-r}$

Now Using the pythagorean theorem on $\triangle BAE$

$r^2 + (h+1)^2 = (r+1)^2$

$r^2 + h^2 + 2h + 1 = r^2 + 2r + 1$

$h^2 + 2h = 2r$

Substituting $h$ in

$(4-4r) + 4\sqrt{1-r} = 2r$

$4\sqrt{1-r} = 6r - 4$

$16-16r = 36r^2 - 48r + 16$

$0 = 36r^2 - 32r$

$r = \frac{32}{36} = \frac{8}{9} \Rightarrow \qquad \textbf{(D)}$

2004 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 19
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2004 AMC 12A Problems/Problem 19

Problem 19

Solution

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