Difference between revisions of "2004 AMC 10A Problems/Problem 23"
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<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> | <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== | + | ==Solution 1== |
<asy> | <asy> | ||
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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>. | ||
+ | ==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>. | ||
== See also == | == See also == |
Revision as of 12:58, 8 February 2016
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 | |
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All AMC 10 Problems and Solutions |
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