A regular octagon $ABCDEFGH$ has side length $1$. Points $X,Y,Z,W$ are the midpoints of $AB,CD,EF,GH$, respectively. What is the area of the intersection of squares $ACEG$ and $XYZW$?

1. Let $O$ be the center and $R=\frac1{2\sin22.5^\circ}$ the circumradius, so $R^2=\frac1{2-\sqrt2}=\frac{2+\sqrt2}2$. The apothem is $a=R\cos22.5^\circ=\frac{1+\sqrt2}2$; $X,Y,Z,W$ are at distance $a$ from $O$, so $[XYZW]=2a^2=\frac{3+2\sqrt2}2$.
2. The sides of $ACEG$ are at distance $\frac R{\sqrt2}$ from $O$. The vertex $X$ is $22.5^\circ$ away from the perpendicular to the nearest side of $ACEG$, so it lies beyond that side by $h=a\cos22.5^\circ-\frac R{\sqrt2}=R\left(\frac{1+\frac{\sqrt2}2}2-\frac{\sqrt2}2\right)=\frac{R(2-\sqrt2)}4$, and $h^2=\frac{2-\sqrt2}{16}$.
3. So each corner of $XYZW$ sticks out of $ACEG$ as a right triangle with angles $22.5^\circ$ and $67.5^\circ$ and altitude $h$ to the hypotenuse; its hypotenuse is $h(\tan22.5^\circ+\tan67.5^\circ)=2\sqrt2\,h$ and its area is $\sqrt2\,h^2=\frac{\sqrt2-1}8$.
4. The overlap is $[XYZW]$ minus four such corners: $\frac{3+2\sqrt2}2-\frac{\sqrt2-1}2=2+\frac{\sqrt2}{2}$.
5. Thus the requested area is $\boxed{2+\frac{\sqrt2}{2}}$.