smc

In any regular polygon with side length $2$, consider the isosceles triangle formed by the center of the polygon $S$ and two consecutive vertices $X$ and $Y$. We are given that $XY=2$. Obviously $SX=SY=r$, where $r$ is the radius of the circumcircle. Let $T$ be the midpoint of $XY$. Then $XT=TY=1$, and $TS=\rho$, where $\rho$ is the radius of the incircle. Applying the Pythagorean theorem on the triangle $STX$, we get that ${\rho }^2+1=r^2$. Then the area between the circumcircle and the incircle can be computed as $\pi r^2-\pi {\rho }^2=\pi r^2-\pi (r^2-1)=\pi$. Hence $A=\pi$, $B=\pi$, and therefore $\boxed{A=B}$.