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Difference between revisions of "2015 AMC 12A Problems/Problem 25"

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==Solution==
 
==Solution==
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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 02:13, 5 February 2015

Problem

A collection of circles in the upper half-plane, all tangent to the $x$-axis, is constructed in layers as follows. Layer $L_0$ consists of two circles of radii $70^2$ and $73^2$ that are externally tangent. For $k>=1$, the circles in $\bigcup_{j=0}^{k-1}L_j$ are ordered according to their points of tangency with the $x$-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 $L_k$ consists of the $2^{k-1}$ circles constructed in this way. Let $S=\bigcup_{j=0}^{6}L_j$, and for every circle $C$ denote by $r(C)$ its radius. What is \[\sum_{C\in S} \frac{1}{\sqrt{r(C)}}?\]

[asy] import olympiad; size(350); defaultpen(linewidth(0.7)); // define a bunch of arrays and starting points pair[] coord = new pair[65]; int[] trav = {32,16,8,4,2,1}; coord[0] = (0,73^2); coord[64] = (2*73*70,70^2); // draw the big circles and the bottom line path arc1 = arc(coord[0],coord[0].y,260,360); path arc2 = arc(coord[64],coord[64].y,175,280); fill((coord[0].x-910,coord[0].y)--arc1--cycle,gray(0.75)); fill((coord[64].x+870,coord[64].y+425)--arc2--cycle,gray(0.75)); draw(arc1^^arc2); draw((-930,0)--(70^2+73^2+850,0)); // We now apply the findCenter function 63 times to get // the location of the centers of all 63 constructed circles. // The complicated array setup ensures that all the circles // will be taken in the right order for(int i = 0;i<=5;i=i+1) { int skip = trav[i]; for(int k=skip;k<=64 - skip; k = k + 2*skip) { pair cent1 = coord[k-skip], cent2 = coord[k+skip]; real r1 = cent1.y, r2 = cent2.y, rn=r1*r2/((sqrt(r1)+sqrt(r2))^2); real shiftx = cent1.x + sqrt(4*r1*rn); coord[k] = (shiftx,rn); } // Draw the remaining 63 circles } for(int i=1;i<=63;i=i+1) { filldraw(circle(coord[i],coord[i].y),gray(0.75)); }[/asy]

$\textbf{(A)}\ \frac{286}{35} \qquad\textbf{(B)}\ \frac{583}{70} \qquad\textbf{(C)}\ \frac{715}{73} \qquad\textbf{(D)}}\ \frac{143}{14} \qquad\textbf{(E)}\ \frac{1573}{146}$ (Error compiling LaTeX. Unknown error_msg)

Solution

Let us start with the two circles in $L_0$ and the circle in $L_1$. Let the larger circle in $L_0$ be named circle $X$ with radius $x$ and the smaller be named circle $Y$ with radius $y$. Also let the single circle in $L_1$ be named circle $Z$ with radius $z$. Draw radii $x$, $y$, and $z$ perpendicular to the x-axis. Drop altitudes $a$ and $b$ from the center of $Z$ to these radii $x$ and $y$, respectively, and drop altitude $c$ from the center of $Y$ to radius $x$ perpendicular to the x-axis. Connect the centers of circles $x$, $y$, and $z$ with their radii, and utilize the Pythagorean Theorem. We attain the following equations. \[(x - z)^2 + a^2 = (x + z)^2 \implies a^2 = 4xz\] \[(y - z)^2 + b^2 = (y + z)^2 \implies b^2 = 4yz\] \[(x - y)^2 + c^2 = (x + y)^2 \implies c^2 = 4xy\]

We see that $a = 2\sqrt{xz}$, $b = 2\sqrt{yz}$, and $c = 2\sqrt{xy}$. Since $a + b = c$, we have that $2\sqrt{xz} + 2\sqrt{yz} = 2\sqrt{xy}$. Divide this equation by $2\sqrt{xyz}$, and this equation becomes the well-known relation of Ford circles $\frac{1}{\sqrt{x}} + \frac{1}{\sqrt{y}} = \frac{1}{\sqrt{z}}.$ We can apply this relationship recursively with the circles in layers $L_2, L_3, \cdots, L_6$.

Here, let $S(n)$ denote the sum of the reciprocals of the square roots of all circles in layer $n$. 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 $\textstyle\sum_{n=0}^{6}S(n)$. We already have that $S(0) = S(1) = \frac{1}{\sqrt{z}} = \frac{1}{73} + \frac{1}{70}$. Then, $S(2) = 2S(1) + S(0) = 3S(0)$. Additionally, $S(3) = 2S(2) + 2S(1) + S(0) = 9S(0)$, and $S(4) = 2S(3) + 2S(2) + 2S(1) + S(0) = 27S(0)$. Now, we notice that, since $S(n)$ is doubled for all $n$ except $0$, $S(n + 1) = 3S(n)$. Hence, our desired sum is $(1 + 1 + 3 + 9 + 27 + 81 + 243)(S(0)) = 365\left(\frac{1}{73} + \frac{1}{70}\right)$. This simplifies to $365\left(\frac{143}{73(70)}\right) = \frac{143}{14} \textbf{ (D)}$.

See Also

2015 AMC 12A (Problems • Answer Key • Resources)
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Problem 24 		Followed by
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All AMC 12 Problems and Solutions
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