$21 \cdot 11$ Given that the side length of square $ABCD$ is 1, and circles with radius 1 are drawn with $A, B, C, D$ as centers, then the area of the common part of the four sectors is
(A) $1-\sqrt{3}+\frac{\pi}{3}$.
(B) $\frac{1}{4}$.
(C) $\frac{\pi}{2}-1$.
(D) $2-\sqrt{3}-\frac{\pi}{8}$.
(China Zhejiang Province Junior High School Mathematics Competition, 1991)
[Solution 1] As shown in Figure (1), let the area of the shaded part be $z$. Then, from
$$
S_{\text {equilateral } \triangle A B P}=\frac{\sqrt{3}}{4}, \text { and } S_{\text {sector } A P D}=\frac{30^{\circ}}{360^{\circ}} \cdot \pi \cdot 1^{2}=\frac{\pi}{12}
$$

we have
$$
\frac{\sqrt{3}}{4}+z+2 \cdot \frac{\pi}{12}=1 \text {, }
$$

which means $z=1-\frac{\sqrt{3}}{4}-\frac{\pi}{6}$.
As shown in Figure (2), let the area of the common part of the four sectors be $x$, and the area of the shaded part be $y$, then
$$
y+2 z=1-\frac{\pi}{4} \text {, i.e., } y=\frac{\pi}{12}+\frac{\sqrt{3}}{2}-1 \text {. }
$$

Thus, $x+4 y+4 z=1$, so $x=1-4 y-4 z$.
Therefore,
$$
x=1-\sqrt{3}+\frac{\pi}{3} \text {. }
$$

Hence, the answer is $(A)$.
[Solution 2] As shown in Figure (2), we can set up the system of equations
$$
\left\{\begin{array}{l}
x+3 y+2 z=\frac{\pi}{4}, \\
y+2 z=1-\frac{\pi}{4}, \\
x+2 y+z=\frac{\pi}{6}+\frac{\pi}{6}-\frac{\sqrt{3}}{4},
\end{array}\right.
$$

Solving this, we get
$$
x=1-\sqrt{3}+\frac{\pi}{3} \text {. }
$$

Hence, the answer is $(A)$.