Example 8. For a regular $n$-sided polygon $(n \geqslant 5)$ centered at point $O$, let two adjacent points be $A, B$. $\triangle X Y Z$ is congruent to $\triangle O A B$. Initially, let $\triangle X Y Z$ coincide with $\triangle O A B$, then move $\triangle X Y Z$ on the plane such that points $Y$ and $Z$ both move along the perimeter of the polygon for one complete revolution, while point $X$ remains inside the polygon. Find the locus of point $X$. (27th IMO)
Solve: Since $\angle Y X Z+$
$$
\begin{array}{l}
\angle Y B Z=\angle A O B+ \\
\angle A B C=\frac{n-2}{n} \pi+\frac{2}{n} \pi \\
=\pi,
\end{array}
$$
$\therefore X, Y, B, Z$ are concyclic, thus $\angle X B Y=$ $\angle X Z Y=\angle O B Y$, so $X$ lies on the extension of $B O$.
$$
\begin{aligned}
\because S_{\triangle X Y B} & =\frac{1}{2} B X \cdot B Y \cdot \sin \angle X Y B \\
& =\frac{1}{2} X Y \cdot B Y \sin \angle X Y B \\
& =\frac{1}{2} X Y \cdot B Y \cdot \sin \angle X Y B, \\
\therefore B X & =\frac{X Y \sin \angle X Y B}{\sin \angle Y B X} \\
& =\frac{X Y \sin \angle X Y B}{\sin \frac{(n-2) \pi}{2 n}} .
\end{aligned}
$$

For the transformation of $X K E$
$$
\frac{(n-2) \pi}{2 n} \leqslant \angle X Y B \leqslant \pi-\frac{(n-2) \pi}{2 n} \text {, }
$$
$\therefore$ the maximum distance from point $X$ to the center $O$ is
$$
d=\frac{X Y}{\sin \left(\frac{\pi}{2}-\frac{\pi}{n}\right)}-O B=\frac{a\left(1-\cos \frac{\pi}{n}\right)}{\sin ^{2} \frac{\pi}{n}},
$$

where $a$ is the side length of the regular $n$-sided polygon. It is evident that as point $Y$ varies on $A B$, point $X$ precisely traces a line segment of length $d$ on the extension of $B O$ from point $O$ twice. Thus, the locus of point $X$ is a "star shape" composed of line segments of length $d$ extending from the center of the regular $n$-sided polygon to each vertex.