2015 AMC 12A Problems/Problem 25
Problem
A collection of circles in the upper half-plane, all tangent to the -axis, is constructed in layers as followsLayer
consists of two circles of radii
and
that are externally tangent. For
, the circles in
are ordered according to their points of tangency with the
-axis. For every pair of consecutive circles in this order, a new circle is constructed externally tangent to each of the two circles in the pair. Layer
consists of the
circles constructed in this way. Let
, and for every circle
denote by
its radius. What is
Solution 1
Let us start with the two circles in and the circle in
. Let the larger circle in
be named circle
with radius
and the smaller be named circle
with radius
. Also let the single circle in
be named circle
with radius
. Draw radii
,
, and
perpendicular to the x-axis. Drop altitudes
and
from the center of
to these radii
and
, respectively, and drop altitude
from the center of
to radius
perpendicular to the x-axis. Connect the centers of circles
,
, and
with their radii, and utilize the Pythagorean Theorem. We attain the following equations.
We see that ,
, and
. Since
, we have that
. Divide this equation by
, and this equation becomes the well-known relation of Descartes's Circle Theorem
We can apply this relationship recursively with the circles in layers
.
Here, let denote the sum of the reciprocals of the square roots of all circles in layer
. The notation in the problem asks us to find the sum of the reciprocals of the square roots of the radii in each circle in this collection, which is
. We already have that
. Then,
. Additionally,
, and
. Now, we notice that
because
, which is a power of
Hence, our desired sum is
. This simplifies to
.
Note that the circles in this question are known as Ford circles.
See Also
2015 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last Problem this year |
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All AMC 12 Problems and Solutions |