jmc

Let $x$ be the length of a leg of the black isosceles triangle. Then the black area is $\frac{1}{2}(x)(x)=\frac{1}{2}{x}^2$. The white area is $6-x^2$. Solving $\frac{1}{2}{x}^2=6-x^2$, we find $x^2=4$, so $x=2$. The distance from A to its original position is the length of a hypotenuse of a right triangle whose legs have length $x$. Therefore, A is $\boxed{2\sqrt{2}}$ centimeters from its original position.