imc

The side length of the inner square traced out by the disk with radius $1$ is $s-4.$ However, there is a piece at each corner (bounded by two line segments and one $90^{\circ }$ arc) where the disk never sweeps out. The combined area of these four pieces is $(1+1{)}^2-\pi \cdot 1^2=4-\pi .$ As a result, we have $$A=s^2-(s-4{)}^2-(4-\pi )=8s-20+\pi .$$ Now, we consider the second disk. The part it sweeps is comprised of four quarter circles with radius $2$ and four rectangles with side lengths of $2$ and $s.$ When we add it all together, we have $2A=8s+4\pi ,$ or $$A=4s+2\pi .$$ We equate the expressions for $A,$ and then solve for $s:$ $$8s-20+\pi =4s+2\pi .$$ We get $s=5+\frac{\pi }{4},$ so the answer is $5+1+4=\boxed{10}.$