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 $k \ge 1$, 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]
$\textbf{(A) } \frac{286}{35} \qquad\textbf{(B) } \frac{583}{70} \qquad\textbf{(C) } \frac{715}{73} \qquad\textbf{(D) } \frac{143}{14} \qquad\textbf{(E) } \frac{1573}{146}$
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 Descartes's Circle Theorem $\frac{1}{\sqrt{x}} + \frac{1}{\sqrt{y}} = \frac{1}{\sqrt{z}}.$ 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 $S(n + 1) = 3S(n)$ because $S(n + 1) = 2S(n) + 2S(n - 1) + \cdots + 2S(1) + S(0)$, which is a power of $3.$ 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 $365\left(\frac{143}{73(70)}\right) = \frac{143}{14} \textbf{(D)}$.
Note that the circles in this question are known as Ford circles.