Difference between revisions of "2001 AMC 12 Problems/Problem 18"
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I Love Math (talk | contribs) m (→Solution 1: Added $S$ to the center of the small circle for easier understanding.) |
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Line 41: | Line 41: | ||
<asy> | <asy> | ||
unitsize(1cm); | unitsize(1cm); | ||
− | pair A=(0,1), B=(4,4), C=(4,1); | + | pair A=(0,1), B=(4,4), C=(4,1), S=(12/9,4/9); |
dot(A); dot(B); | dot(A); dot(B); | ||
draw( circle(A,1) ); | draw( circle(A,1) ); | ||
Line 51: | Line 51: | ||
label("$B$",B,N); | label("$B$",B,N); | ||
label("$C$",C,E); | label("$C$",C,E); | ||
+ | label("$S$",S,N); | ||
− | filldraw( circle( | + | filldraw( circle(S,4/9), lightgray, black ); |
− | dot( | + | dot(S); |
draw( rightanglemark(A,C,B) ); | draw( rightanglemark(A,C,B) ); | ||
+ | draw( S -- A ); | ||
+ | draw( S -- B ); | ||
</asy> | </asy> | ||
Revision as of 22:43, 2 March 2019
Contents
[hide]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
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.