AMC 12 2015a Problem 25

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 $k\ge1$, 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 $$\sum_{C\in S} \frac{1}{\sqrt{r(C)}}?$$

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));
}

\textbf{(A)}\ \frac{286}{35} \textbf{(B)}\ \frac{583}{70} \textbf{(C)}\ \frac{715}{73} \textbf{(D)}\ \frac{143}{14} \textbf{(E)}\ \frac{1573}{146}