2001 AMC 12 Problems/Problem 18
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 .
Let be the radius of the small circle, and let be the perpendicular distance from to . 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 , which puts the center on the wrong side of , so this is not the correct case.
(Note: This case corresponds to the other circle that is tangent to both given circles and the common tangent line. By coincidence, due to the ratio between radii of and , this circle turns out to have the same radius as circle , with center directly left of center , and tangent to directly above center .)
The second case solves to . We then have , hence .
More generally, for two large circles of radius and , the radius of the small circle is .
Equivalently, we have that .
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 .
Video Solution
See Also
2001 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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