Difference between revisions of "2015 AMC 12A Problems/Problem 25"
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==Solution== | ==Solution== | ||
+ | Let us start with the two circles in <math>L_0</math> and the circle in <math>L_1</math>. Let the larger circle in <math>L_0</math> be named circle <math>X</math> with radius <math>x</math> and the smaller be named circle <math>Y</math> with radius <math>y</math>. Also let the single circle in <math>L_1</math> be named circle <math>Z</math> with radius <math>z</math>. Draw radii <math>x</math>, <math>y</math>, and <math>z</math> perpendicular to the x-axis. Drop altitudes <math>a</math> and <math>b</math> from the center of <math>Z</math> to these radii <math>x</math> and <math>y</math>, respectively, and drop altitude <math>c</math> from the center of <math>Y</math> to radius <math>x</math> perpendicular to the x-axis. Connect the centers of circles <math>x</math>, <math>y</math>, and <math>z</math> with their radii, and utilize the Pythagorean Theorem. We attain the following equations. <cmath>(x - z)^2 + a^2 = (x + z)^2 \implies a^2 = 4xz</cmath> <cmath>(y - z)^2 + b^2 = (y + z)^2 \implies b^2 = 4yz</cmath> <cmath>(x - y)^2 + c^2 = (x + y)^2 \implies c^2 = 4xy</cmath> | ||
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+ | We see that <math>a = 2\sqrt{xz}</math>, <math>b = 2\sqrt{yz}</math>, and <math>c = 2\sqrt{xy}</math>. Since <math>a + b = c</math>, we have that <math>2\sqrt{xz} + 2\sqrt{yz} = 2\sqrt{xy}</math>. Divide this equation by <math>2\sqrt{xyz}</math>, and this equation becomes the well-known relation of Ford circles <math>\frac{1}{\sqrt{x}} + \frac{1}{\sqrt{y}} = \frac{1}{\sqrt{z}}.</math> We can apply this relationship recursively with the circles in layers <math>L_2, L_3, \cdots, L_6</math>. | ||
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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, since <math>S(n)</math> is doubled for all <math>n</math> except <math>0</math>, <math>S(n + 1) = 3S(n)</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>. | ||
== See Also == | == See Also == | ||
{{AMC12 box|year=2015|ab=A|num-b=24|after=Last Problem}} | {{AMC12 box|year=2015|ab=A|num-b=24|after=Last Problem}} |
Revision as of 03:13, 5 February 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
$\textbf{(A)}\ \frac{286}{35} \qquad\textbf{(B)}\ \frac{583}{70} \qquad\textbf{(C)}\ \frac{715}{73} \qquad\textbf{(D)}}\ \frac{143}{14} \qquad\textbf{(E)}\ \frac{1573}{146}$ (Error compiling LaTeX. ! Extra }, or forgotten $.)
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 Ford circles 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, since is doubled for all except , . Hence, our desired sum is . This simplifies to .
See Also
2015 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last Problem |
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 |