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 $5^2$ and $8^2$ 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 $S=\bigcup_{j=1}^{4}L_j$, and for every circle $C$ denote by $r(C)$ its radius. What is $\sum_{C\in S}\frac1{\sqrt{r(C)}}$?

1. Suppose two circles of radii $r_1,r_2$ are externally tangent and both tangent to the $x$-axis. The distance between their points of tangency with the axis is $\sqrt{(r_1+r_2)^2-(r_1-r_2)^2}=2\sqrt{r_1r_2}$. If a third circle of radius $r$ is tangent to both and to the axis between them, then $2\sqrt{r_1r}+2\sqrt{rr_2}=2\sqrt{r_1r_2}$, which gives $\dfrac1{\sqrt r}=\dfrac1{\sqrt{r_1}}+\dfrac1{\sqrt{r_2}}$.
2. Write $\rho(C)=\dfrac1{\sqrt{r(C)}}$. By (i), each new circle has $\rho$ equal to the sum of the $\rho$-values of the two circles it was built on. In $L_0$ the values are $a=\frac15$ and $b=\frac18$, and we set $s=a+b=\frac{13}{40}$.
3. Let $T_k$ be the sum of $\rho$ over layer $L_k$, so $T_0=s$. For $k\ge1$ the new circles correspond to the pairs of consecutive circles in the ordered list of layers $L_0,\dots,L_{k-1}$. Every circle in that list lies in two such pairs, except the two end circles (with values $a$ and $b$), which lie in one. Hence $T_k=2(T_0+\dots+T_{k-1})-s$.
4. Compute: $T_1=2s-s=s$; $T_2=2(2s)-s=3s$; $T_3=2(5s)-s=9s$; $T_4=2(14s)-s=27s$.
5. So $\displaystyle\sum_{C\in S}\frac1{\sqrt{r(C)}}=T_1+T_2+T_3+T_4=(1+3+9+27)s=40\cdot\frac{13}{40}=\boxed{13}$.