Difference between revisions of "2000 AMC 12 Problems/Problem 24"
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<cmath>\left(\frac {r_1}{2}\right)^2 + r_2^2 = (r_1 - r_2)^2</cmath> | <cmath>\left(\frac {r_1}{2}\right)^2 + r_2^2 = (r_1 - r_2)^2</cmath> | ||
− | After simplification, <math>r_2 = \frac{3r_1}{8}</math>. | + | After simplification, <math>r_2 = \frac{3r_1}{8}</math>. |
== See also == | == See also == |
Revision as of 17:12, 21 November 2017
Problem
If circular arcs and
have centers at
and
, respectively, then there exists a circle tangent to both
and
, and to
. If the length of
is
, then the circumference of the circle is
Solution
Since are all radii, it follows that
is an equilateral triangle.
Draw the circle with center and radius
. Then let
be the point of tangency of the two circles, and
be the intersection of the smaller circle and
. Let
be the intersection of the smaller circle and
. Also define the radii
(note that
is a diameter of the smaller circle, as
is the point of tangency of both circles, the radii of a circle is perpendicular to the tangent, hence the two centers of the circle are collinear with each other and
).
By the Power of a Point Theorem,
Since , then
. Since
is equilateral,
, and so
. Thus
and the circumference of the circle is
.
(Alternatively, the Pythagorean Theorem can also be used to find in terms of
. Notice that since AB is tangent to circle
,
is perpendicular to
. Therefore,
After simplification, .
See also
2000 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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.