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

<center>[[Image:AMC10_2004A_23.png]]</center>

<math> \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} </math>

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

Also, right triangle <math>A_1B_1E</math> has legs <math>r, 1+x</math>, and hypotenuse <math>1+r</math>. Solving, 

<cmath>\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*}</cmath> 
 
So the answer is <math>\mathrm{(D)}</math>.

== See also ==
* <url>viewtopic.php?t=131335 AoPS topic</url>
{{AMC10 box|year=2004|ab=A|num-b=22|num-a=24}}

[[Category:Introductory Geometry Problems]]