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. The radius of the third circle is
Solution
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
.
See Also
2001 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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