2001 AMC 12 Problems/Problem 18
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. The radius of the third circle is
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 .
The horizontal line is the equivalent of a circle of curvature , thus we can apply Descartes' Circle Formula.
The four circles have curvatures , and .
Simplifying, we get
Obviously cannot equal , therefore .
|2001 AMC 12 (Problems • Answer Key • Resources)|
|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.