Difference between revisions of "2015 AMC 12A Problems/Problem 25"
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Here, let <math>S(n)</math> denote the sum of the reciprocals of the square roots of all circles in layer <math>n</math>. 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 <math>\textstyle\sum_{n=0}^{6}S(n)</math>. We already have that <math>S(0) = S(1) = \frac{1}{\sqrt{z}} = \frac{1}{73} + \frac{1}{70}</math>. Then, <math>S(2) = 2S(1) + S(0) = 3S(0)</math>. Additionally, <math>S(3) = 2S(2) + 2S(1) + S(0) = 9S(0)</math>, and <math>S(4) = 2S(3) + 2S(2) + 2S(1) + S(0) = 27S(0)</math>. Now, we notice that <math>S(n + 1) = 3S(n)</math> because <math>S(n + 1) = 2S(n) + 2S(n - 1) + \cdots + 2S(1) + S(0)</math>, which is a power of <math>3.</math> Hence, our desired sum is <math>(1 + 1 + 3 + 9 + 27 + 81 + 243)(S(0)) = 365\left(\frac{1}{73} + \frac{1}{70}\right)</math>. This simplifies to <math>365\left(\frac{143}{73(70)}\right) = \frac{143}{14} \textbf{ (D)}</math>. | Here, let <math>S(n)</math> denote the sum of the reciprocals of the square roots of all circles in layer <math>n</math>. 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 <math>\textstyle\sum_{n=0}^{6}S(n)</math>. We already have that <math>S(0) = S(1) = \frac{1}{\sqrt{z}} = \frac{1}{73} + \frac{1}{70}</math>. Then, <math>S(2) = 2S(1) + S(0) = 3S(0)</math>. Additionally, <math>S(3) = 2S(2) + 2S(1) + S(0) = 9S(0)</math>, and <math>S(4) = 2S(3) + 2S(2) + 2S(1) + S(0) = 27S(0)</math>. Now, we notice that <math>S(n + 1) = 3S(n)</math> because <math>S(n + 1) = 2S(n) + 2S(n - 1) + \cdots + 2S(1) + S(0)</math>, which is a power of <math>3.</math> Hence, our desired sum is <math>(1 + 1 + 3 + 9 + 27 + 81 + 243)(S(0)) = 365\left(\frac{1}{73} + \frac{1}{70}\right)</math>. This simplifies to <math>365\left(\frac{143}{73(70)}\right) = \frac{143}{14} \textbf{ (D)}</math>. | ||
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Revision as of 20:00, 16 September 2015
Problem
A collection of circles in the upper half-plane, all tangent to the -axis, is constructed in layers as follows. Layer
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
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
.