Difference between revisions of "2004 AMC 10A Problems/Problem 23"
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So the answer is <math>\boxed{\mathrm{(D)}\ \frac{8}{9}}</math>. | So the answer is <math>\boxed{\mathrm{(D)}\ \frac{8}{9}}</math>. | ||
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==Solution 2== | ==Solution 2== | ||
Using [[Descartes' Circle Formula]], <math>\left(1-\frac{1}{2}+\frac{1}{r}+\frac{1}{r}\right)^2=2\left(\frac{1}{4}+1+\frac{1}{r^2}+\frac{1}{r^2}\right)</math>. Solving this gives us linear equation with <math>r=\boxed{\mathrm{(D)}\ \frac{8}{9}}</math>. | Using [[Descartes' Circle Formula]], <math>\left(1-\frac{1}{2}+\frac{1}{r}+\frac{1}{r}\right)^2=2\left(\frac{1}{4}+1+\frac{1}{r^2}+\frac{1}{r^2}\right)</math>. Solving this gives us linear equation with <math>r=\boxed{\mathrm{(D)}\ \frac{8}{9}}</math>. |
Revision as of 18:32, 5 November 2017
Contents
[hide]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 1
Let be the center of circle for all and let be the tangent point of . Since the radius of is the diameter of , the radius of is . Let the radius of be and let . If we connect , we get an isosceles triangle with lengths . Then right triangle has legs and hypotenuse . Solving for , we get .
Also, right triangle has legs , and hypotenuse . Solving,
So the answer is .
Solution 2
Using Descartes' Circle Formula, . Solving this gives us linear equation with .
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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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