Three circles are drawn in the first quadrant of the $xy$-plane. The first circle has radius $2s$ and is tangent to both axes. The other two circles each have radius $s$: the second is tangent to the first circle and the $x$-axis, and the third is tangent to the first circle and the $y$-axis. A circle of radius $r > 2s$ is tangent to both axes and to the second and third circles. What is $r/s$?

1. Place the axes as in the figure. The first circle has radius $2s$, so its center is $C=(2s,2s)$. The second circle has radius $s$ and is tangent to the $x$-axis, so its center is $D=(x,s)$. It is tangent to the first circle, so $CD=2s+s=3s$: $(x-2s)^2+(s-2s)^2=9s^2$, so $x-2s=2\sqrt2\,s$ and $D=\left((2+2\sqrt2)s,\,s\right)$. (By symmetry the third circle has center $\left(s,(2+2\sqrt2)s\right)$.)
2. The big circle has center $E=(r,r)$ and is tangent to the second circle, so $DE=r+s$: $$\left(r-(2+2\sqrt2)s\right)^2+(r-s)^2=(r+s)^2.$$
3. Expanding and simplifying gives $r^2-(8+4\sqrt2)sr+(12+8\sqrt2)s^2=0$, which factors as $(r-2s)\left(r-(6+4\sqrt2)s\right)=0$.
4. The root $r=2s$ is the first circle itself, so $r>2s$ forces $r=(6+4\sqrt2)s$, and $\dfrac rs=\boxed{6+4\sqrt2}\approx11.66$.