$ABCD$ is a square with $AB=\sqrt2$. Point $P$ is on diagonal $\overline{AC}$ such that $AP=2-\sqrt2$. The square region is rotated $90^\circ$ counterclockwise about $P$, sweeping out a region $S$. What is the area of $S$?
Put $p=AP/\sqrt2=\sqrt2-1$ and $q=AB-p=1$. Let $\beta=\arctan(p/q)=22.5^\circ$.
As in the problem, the swept region consists of four sectors centered at $P$ (the quarter circles swept by $A$ and $C$, and sectors of angle $2\beta$ swept by $B$ and $D$) and four triangles. The sectors have total area $(p^2+q^2)(\frac\pi2+2\beta)$, and the triangles have total area $q^2-p^2$.
