Difference between revisions of "2001 AMC 12 Problems/Problem 18"
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== Problem == | == Problem == | ||
− | A circle centered at <math>A</math> with a radius of 1 and a circle centered at <math>B</math> with a radius of 4 are externally tangent. A third circle is tangent to the first two and to one of their common external tangents as shown. | + | A circle centered at <math>A</math> with a radius of 1 and a circle centered at <math>B</math> with a radius of 4 are externally tangent. A third circle is tangent to the first two and to one of their common external tangents as shown. What is the radius of the third circle? |
<asy> | <asy> |
Latest revision as of 15:15, 17 September 2020
Problem
A circle centered at with a radius of 1 and a circle centered at with a radius of 4 are externally tangent. A third circle is tangent to the first two and to one of their common external tangents as shown. What is the radius of the third circle?
Solution
Solution 1
In the triangle we have and , thus by the Pythagorean theorem we have .
We can now pick a coordinate system where the common tangent is the axis and lies on the axis. In this coordinate system we have and .
Let be the radius of the small circle, and let be the -coordinate of its center . We then know that , as the circle is tangent to the axis. Moreover, the small circle is tangent to both other circles, hence we have and .
We have and . Hence we get the following two equations:
Simplifying both, we get
As in our case both and are positive, we can divide the second one by the first one to get .
Now there are two possibilities: either , or . In the first case clearly , hence this is not the correct case. (Note: This case corresponds to the other circle that is tangent to both given circles and the axis - a large circle whose center is somewhere to the left of .) The second case solves to . We then have , hence .
Solution 2
The horizontal line is the equivalent of a circle of curvature , thus we can apply Descartes' Circle Formula.
The four circles have curvatures , and .
We have
Simplifying, we get
Obviously cannot equal , therefore .
See Also
2001 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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.