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Difference between revisions of "2004 AMC 10A Problems/Problem 23"

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==Solution==
 
==Solution==
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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>.  
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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, 2r</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,  
 
Also, right triangle <math>A_1B_1E</math> has legs <math>r, 1+x</math>, and hypotenuse <math>1+r</math>. Solving,  

Revision as of 09:10, 1 May 2011

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 $B$?

AMC10 2004A 23.png

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

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 $1 + r, 2r$. 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 $x^2 = (2-r)^2 - r^2 \Longrightarow x = \sqrt{4-4r}$ (Error compiling LaTeX. Unknown error_msg).

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 $\mathrm{(D)}$.

See also

  • <url>viewtopic.php?t=131335 AoPS topic</url>
2004 AMC 10A (Problems • Answer Key • Resources)
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Problem 24
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All AMC 10 Problems and Solutions
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