Difference between revisions of "2001 AMC 12 Problems/Problem 18"
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− | More generally, for two large circles of radius <math>a</math> and <math>b</math>, the radius of the small circle is <math>\frac{ab}{(\sqrt{a}+\sqrt{b})^2}</math> | + | More generally, for two large circles of radius <math>a</math> and <math>b</math>, the radius of the small circle is <math>\frac{ab}{\left(\sqrt{a}+\sqrt{b}\right)^2}</math> |
=== Solution 2 === | === Solution 2 === |
Revision as of 00:01, 20 August 2023
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
, hence 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 - a large circle whose center is somewhere to the left of
.) 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
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 |
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 |
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