Difference between revisions of "2004 AMC 10A Problems/Problem 23"
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==Solution== | ==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 == | == See also == | ||
* <url>viewtopic.php?t=131335 AoPS topic</url> | * <url>viewtopic.php?t=131335 AoPS topic</url> | ||
{{AMC10 box|year=2004|ab=A|num-b=22|num-a=24}} | {{AMC10 box|year=2004|ab=A|num-b=22|num-a=24}} | ||
+ | |||
+ | [[Category:Introductory Geometry Problems]] |
Revision as of 15:35, 16 October 2007
Problem
Circles , , and are externally tangent to each other and internally tangent to circle . Circles and are congruent. Circle has radius and passes through the center of . What is the radius of circle ?
Solution
Let be the center of , and be the intersection point of . Since the radius of is the diameter of , the radius of is . Let the radius of be . If we connect the centers of the circles (we will denote these as , we get an isosceles triangle with lengths . Also, is the difference between the radius of , , and , so right has legs and hypotenuse . Solving for , we get $x^2 = (2-r)^2 - r^2 \Longrightarow x = \sqrt{4-4r}$ (Error compiling LaTeX. Unknown error_msg).
Also, right triangle has legs , and hypotenuse . Solving,
So the answer is .
See also
- <url>viewtopic.php?t=131335 AoPS topic</url>
2004 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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All AMC 10 Problems and Solutions |