Question: Let a regular $n$-sided polygon with side length 1 in the plane, whose vertices are $P_{1}, P_{2}, \cdots, P_{n}$. If two different points $P_{n+1} 、 P_{n+2}$ are placed arbitrarily within the shape or on its boundary, try to find: $\min _{\leqslant i<j \leqslant n+2} P_{i} P_{j}$'s maximum value.
Solution: Take points $P_{n+1}$ and $P_{n+2}$ on the perpendicular bisectors of $P_{1} P_{2}$ and $P_{\left[\frac{n}{2}\right]+1} P_{\left[\frac{n}{2}\right]+2}$, respectively, such that $P_{n+1}$ and $P_{n+2}$ are inside the $n$-sided polygon, and
$P_{1} P_{n+1}=P_{n+1} P_{n+2}=P_{n+2} P_{\left[\frac{n}{2}\right]+1}$
We will discuss this in two cases.
(1) If $n$ is even, let $n=2 k\left(k \in \mathbf{N}_{+}\right)$, and

let point $O$ be the center of the regular $n$-sided polygon, $r=O P_{i}(i=1,2, \cdots, n)$, then we have
$2 r \sin \frac{\pi}{n}=1$.
Thus, the distance between $P_{1} P_{2}$ and $P_{k+1} P_{k+2}$ is
$2 r \sin \frac{(k-1) \pi}{n}=\frac{\sin \frac{(k-1) \pi}{n}}{\sin \frac{\pi}{n}}=\frac{\cos \frac{\pi}{n}}{\sin \frac{\pi}{n}}$.
As shown in Figure 1, let $P_{1} P_{n+1}=P_{n+1} P_{n+2}=P_{n+2} P_{k+2}$
$=x$. Then we have
$x+2 \sqrt{x^{2}-\frac{1}{4}}$
$=\frac{\cos \frac{\pi}{n}}{\sin \frac{\pi}{n}}$.
Let $t=\frac{\cos \frac{\pi}{n}}{\sin \frac{\pi}{n}}$.
Then the original equation can be transformed into
$\sqrt{4 x^{2}-1}=t-x$. Squaring both sides,
$3 x^{2}+2 t x-\left(t^{2}+1\right)=0$
Solving for $x$, we get $x=\frac{-2 t+\sqrt{4 t^{2}+12\left(t^{2}+1\right)}}{6}$ (negative value

discarded), i.e., $\square$
$x=\frac{\sqrt{4 t^{2}+3}-t}{3}$
Substituting $t=\frac{\cos \frac{\pi}{n}}{\sin \frac{\pi}{n}}$ into the above equation, we get
$x=\frac{\sqrt{3+\cos ^{2} \frac{\pi}{n}}-\cos \frac{\pi}{n}}{3 \sin \frac{\pi}{n}}$.
At this point, $\min _{1 \leqslant i8$ when, obviously, the distance between adjacent vertices

satisfies the condition, so the maximum value at this time is 1.
Next, we prove: Equation (1) is the maximum value among the minimum values of $P_{i} P_{i}$.

By contradiction: If there exists a method such that the minimum value of the distances from $P_{n+1}$ and $P_{n+2}$ to each vertex and $P_{n+1} P_{n+2}$ is achieved by drawing arcs inside the $n$-sided polygon, then, inside the $n$-sided polygon (in the middle), we get a curved $n$-sided polygon $A_{1} A_{2} \cdots A_{1}$
Since $P_{n+1}$ and $P_{n+2}$ are inside the curved $n$-sided polygon $A_{1} A_{2} \cdots A_{n}$, $P_{n+1} P_{n+2}$ is also inside the regular $n$-sided polygon $A_{1} A_{2} \cdots A_{n}$. This leads to $P_{n+1} P_{n+2}8$ can be uniformly addressed, then enumerate $n$, and finally summarize and generalize.