2004 AMC 12A Problems/Problem 19
- The following problem is from both the 2004 AMC 12A #19 and 2004 AMC 10A #23, so both problems redirect to this page.
Problem 19
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
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
Note that since is the center of the larger circle of radius . Using the Pythagorean Theorem on ,
Now using the Pythagorean Theorem on ,
Substituting ,
Solution 3
We can apply Descartes' Circle Formula.
The four circles have curvatures , and .
We have
Simplifying, we get
See Also
2004 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.