Given a unit circle $\odot O$. A side $AB$ of the square $ABCD$ is a chord of $\odot O$. Try to find $OD_{\max}$ and $OD_{\min}$.
Solve as shown in Figure 2, draw $O M \perp A B$ at point $M$, and let
$$
\angle O A B=\alpha \text {. }
$$

Then $A D=A B=2 A M=2 \cos \alpha$.
By the cosine rule, we have
$$
\begin{array}{l} 
O D^{2}= 1^{2}+(2 \cos \alpha)^{2}- \\
2 \times 1(2 \cos \alpha) \cos \left(90^{\circ}+\alpha\right) \\
= 1+4 \cos ^{2} \alpha+4 \sin \alpha \cdot \cos \alpha \\
= 1+4 \cos ^{2} \alpha+2 \sin 2 \alpha \\
= 1+2(\cos 2 \alpha+1)+2 \sin 2 \alpha \\
= 2(\sin 2 \alpha+\cos 2 \alpha)+3 \\
= 2 \sqrt{2} \sin \left(2 \alpha+45^{\circ}\right)+3 \\
\leqslant 2 \sqrt{2}+3=(\sqrt{2}+1)^{2} \\
\Rightarrow O D \leqslant \sqrt{2}+1 . \\
\Rightarrow O D_{\max }=\sqrt{2}+1 .
\end{array}
$$

It is easy to see that $A D=2 D G=2 O K$.
Let $\angle O A K=\alpha$. Then $A D=2 \sin \alpha$.
By the cosine rule, we have
$$
\begin{array}{l}
O D^{2}=3-2 \times \sqrt{2} \sin \left(2 \alpha+45^{\circ}\right) \\
\geqslant 3-2 \sqrt{2}=(\sqrt{2}-1)^{2} \\
\Rightarrow O D \geqslant \sqrt{2}-1 \\
\Rightarrow O D_{\text {min }}=\sqrt{2}-1 .
\end{array}
$$
(Huang Quanfu, Jiangzhen Middle School, Huaining County, Anhui Province, 246142)
As shown in Figure 3, draw $O G \perp C D$ at point $G, O K \perp A D$ at point $K$.