A disk of radius $2$ rolls all the way around the inside of a square of side length $s>8$ and sweeps out a region of area $A$. A second disk of radius $2$ rolls all the way around the outside of the same square and sweeps out a region of area $\frac32A$. The value of $s$ can be written as $a+\frac{b\pi}{c}$, where $a,b$, and $c$ are positive integers and $b$ and $c$ are relatively prime. What is $a+b+c$?

1. Inside: the disk (radius $2$) sweeps the square minus the unswept central square of side $s-8$ and minus four corner pieces, each of area $2^2-\frac\pi4\cdot2^2=4-\pi$. So $A=s^2-(s-8)^2-4(4-\pi)=16s-80+4\pi$.
2. Outside: the swept region consists of four rectangles, each $s\times4$, and four quarter-disks of radius $4$ at the corners, which together make one disk of radius $4$. So the swept area is $16s+16\pi$.
3. Setting $16s+16\pi=\frac32(16s-80+4\pi)=24s-120+6\pi$ gives $8s=120+10\pi$, so $s=15+\dfrac{5\pi}4$ (which indeed exceeds $8$).
4. $a=15,b=5,c=4$, so $a+b+c=\boxed{24}$.