A circle $O$ with radius $x$ is tangent to at least one side of a square with side length 1, and rolls inside it. After rolling around the square once, let the area of the region not covered by circle $O$ be $S$. The value of $x$ that minimizes $S$ is
(A) $4-\pi$
(B) $\frac{1}{4}$
(C) $\frac{4}{20-\pi}$
(D) $\frac{1}{2}$
5. (C).

When $0 \leqslant x<\frac{1}{4}$, the area $S$ of the 4 right angles and the central small square of the positive fractal is given by
$$
S=(20-\pi)\left(x-\frac{4}{20-\pi}\right)^{2}+\frac{4-\pi}{20-\pi} .
$$

When $\frac{1}{4} \leqslant x \leqslant \frac{1}{2}$, the area $S$ of the 4 corner regions is given by $S=(4-\pi) x^{2}$.
Thus, when $x=\frac{4}{20-\pi}$, $S$ achieves its minimum value, which is $\frac{4-\pi}{20-\pi}$.