[Geometry]
Problem: Points $P, Q, R$, and $S$ lie in the interior of square $A B C D$ such that triangles $A B P, B C Q$, $C D R$, and $D A S$ are equilateral. If $A B=1$, compute the area of quadrilateral $P Q R S$.

Solution: $P Q R S$ is a square with diagonal $\overline{R P}$. Extend $\overline{R P}$ to intersect $\overline{A B}$ and $\overline{C D}$ at $M$ and $N$ respectively, as shown in the diagram below.

<img_3457>

Then $\overline{M P}$ is an altitude of $\triangle A B P$ and $\overline{R N}$ is an altitude of $\triangle C D R$. Adding lengths, $M P+R N=M R+2 R P+P N=1+R P$, so $R P=\sqrt{3}-1$. Therefore $[P Q R S]=\frac{1}{2}(R P)^{2}=$ $2-\sqrt{3}$.

Final answer: $2-\sqrt{3}$