2004 AMC 10A Problems/Problem 23

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Problem

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 $B$ ?

$\mathrm{(A) \ } \frac{2}{3} \qquad \mathrm{(B) \ } \frac{\sqrt{3}}{2} \qquad \mathrm{(C) \ } \frac{7}{8} \qquad \mathrm{(D) \ } \frac{8}{9} \qquad \mathrm{(E) \ } \frac{1+\sqrt{3}}{3}$

Solution

Let $O$ be the center of $D$ , and $E$ be the intersection 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 $r$ . If we connect the centers of the circles $A, B, C$ (we will denote these as $A_1, B_1, C_1$ , we get an isosceles triangle with lengths $1 + r, r$ . Also, $B_1E$ is the difference between the radius of $D$ , $2$ , and $r$ , so right $\triangle OB_1E$ has legs $r, x$ and hypotenuse $2-r$ . Solving for $x$ , we get $x^2 = (2-r)^2 - r^2 \Longrightarow x = \sqrt{4-4r}$ .

Also, right triangle $A_1B_1E$ 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 $\mathrm{(D)}$ .

2004 AMC 10A (Problems)
Preceded by Problem 22	Followed by Problem 24
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2004 AMC 10A Problems/Problem 23

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