Example 2 If a square with side length 1 can be covered by 3 circular paper pieces, all with diameter equal to $a$, find the minimum value of $a$.


Let $A B C D$ be a square with side length 1, and $E, F$ be points on $A D$ and $B C$ respectively, such that $D E = C F = \frac{7}{8}$. Also, $G, H$ are the midpoints of $C D$ and $E F$ respectively, and connect $G H$ (as shown in Figure 6-23). Thus, $D H = C H = \sqrt{\left(\frac{1}{2}\right)^{2} + \left(\frac{7}{8}\right)^{2}} = \frac{\sqrt{65}}{8}$, and $A F = \sqrt{\left(\frac{1}{8}\right)^{2} + 1^{2}} = \frac{\sqrt{65}}{8}$.

The circles with diameters $D H, C H, A F$ cover the square $A B C D$. Therefore, the square can be covered by 3 circular paper pieces with diameters all equal to $\frac{\sqrt{65}}{8}$, so the minimum value of $a$ is $\leqslant \frac{\sqrt{65}}{8}$.

If the square $A B C D$ can be covered by 3 circles with radii all equal to $b (b < \frac{\sqrt{65}}{16})$, then the centers of these circles, denoted as $M_{1}, M_{2}, M_{3}$, must be within the square. Since $E$ is not covered by $M_{1}$, assume $E$ is covered by $M_{2}$. Because $G E = \frac{\sqrt{65}}{8} > b$ and $G B > G F = \frac{\sqrt{65}}{8} > b$, $G$ cannot be covered by $M_{1}$ or $M_{2}$, and can only be covered by $M_{3}$. Since $F A = F G = \frac{\sqrt{65}}{8} > b$, $F$ cannot be covered by $M_{1}$ or $M_{3}$, and can only be covered by $M_{2}$. Take a point $K$ on $F C$ such that $F K = \frac{1}{8}$, then $K A > K E = \frac{\sqrt{65}}{8} > b$, so $K$ cannot be covered by $M_{1}$ or $M_{2}$, and can only be covered by $M_{3}$. Since $D B > D F > D K > \frac{\sqrt{65}}{8} > b$, and $B, F, K$ are in $M_{1}, M_{2}, M_{3}$ respectively, $D$ cannot be covered by $M_{1}, M_{2}$, or $M_{3}$, which contradicts the assumption that $M_{1}, M_{2}, M_{3}$ can cover the square $A B C D$. Therefore, the square $A B C D$ cannot be covered by 3 equal circles with diameters less than $\frac{\sqrt{65}}{8}$.

In conclusion, the minimum value of $a$ is $\frac{\sqrt{65}}{8}$.