Octagon Formed by Cutting Corner of Square

Let $x$ be the length cut off from each corner along each edge of the square. The hypotenuse of the corner right triangle (side of the regular octagon) is $\sqrt{x^2+x^2} = x\sqrt{2}$. The remaining middle part on each side of the original square is $2 - 2x$. Since all sides of the octagon are equal: $x\sqrt{2} = 2 - 2x \implies x(\sqrt{2}+2) = 2 \implies x = \frac{2}{2+\sqrt{2}}$. The side length of the octagon is $x\sqrt{2} = \frac{2\sqrt{2}}{\sqrt{2}(\sqrt{2}+1)} = \frac{2}{\sqrt{2}+1}$.