5. A square $A B C D$ with side 1 is inscribed in a circle $\gamma$. Construct the symmetrics of the arcs $\overparen{A B}, \overparen{B C}, \overparen{C D}, \overparen{D A}$ of $\gamma$ with respect to the sides $A B, B C, C D$, $D A$ respectively. Let $L, M, N, O$ be the midpoints of the arcs thus obtained; what is the area of $L M N O$?
(A) $\frac{\sqrt{2}}{4}$
(B) $\sqrt{2}-1$
(C) $1-\frac{\sqrt{2}}{2}$
(D) $\frac{1}{4}$
(E) $3-2 \sqrt{2}$.

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5. The answer is (E). The radius of the circle circumscribing the square is $\frac{\sqrt{2}}{2}$, being half the diagonal of the square itself. $L M N O$ is also a square, being a figure symmetric under 90-degree rotations. Now, the vertices of the unit square divide the circle into four arcs. The midpoints of these arcs have the same distance from the sides of the square as the points $L, M, N, O$, being symmetric to them with respect to the respective sides. This distance is the difference between the radius of the circle and half the side of the square, thus $\frac{\sqrt{2}-1}{2}$. $L N$ is equal to the difference between the side of the unit square and twice the above distance, which is $2-\sqrt{2}$. Finally, $L N$ is the diagonal of $L M N O$, whose side is therefore $\sqrt{2}-1$, and whose area is $(\sqrt{2}-1)^{2}=3-2 \sqrt{2}$.