Example 10 As shown in Figure 1.10.4, for a regular $n(n \geqslant 5)$-sided polygon centered at point $O$, two adjacent vertices are denoted as $A, B$, and $\triangle X Y Z$ is congruent to $\triangle O A B$. Initially, let $\triangle X Y Z$ overlap with $\triangle O A B$, then move $X Y Z$ on the plane so that points $Y$ and $Z$ both move along the perimeter of the polygon for one complete round, while point $X$ remains moving inside the polygon. Find the trajectory of point $X$.
Let $O$ be the origin to establish a complex plane. Suppose point $C$ is another vertex of the regular $n$-sided polygon adjacent to point $B$, and $A, B, C$ are in counterclockwise order. Let at a certain moment during the movement of $\triangle XYZ$, $Y$ is on $AB$ and $Z$ is on $BC$. According to the section formula, we have
$$
\begin{array}{ll}
\vec{Y}=(1-\lambda) \vec{A}+\lambda \vec{B} & (0 \leqslant \lambda \leqslant 1) \\
\vec{Z}=(1-\mu) \vec{B}+\mu \vec{C} & (0 \leqslant \mu \leqslant 1)
\end{array}
$$

Let $\varepsilon=\mathrm{e}^{\frac{i 2 \pi}{n}}$, then $\vec{B}=\vec{\varepsilon}, \vec{C}=\vec{\varepsilon}=\varepsilon^{2} \vec{A}$, and $(\vec{Y}-\vec{X}) \varepsilon=\vec{Z}-\vec{X}$. Therefore,
$$
\vec{X}=\frac{\vec{Z}-\vec{Y} \varepsilon}{1-\varepsilon}=\frac{1}{1-\varepsilon}(\mu-\lambda)(\vec{C}-\vec{B})=\frac{1}{1-\varepsilon}(\mu-\lambda)(\varepsilon \vec{B}-\vec{B})
$$

That is,
$$
\vec{X}=-(\mu-\lambda) \vec{B}
$$

Since $\mu-\lambda$ is a real number, the above equation indicates that point $X$ always lies on the line determined by $O$ and $B$. Next, from
$$
|\vec{A}|=|\vec{Y}-\vec{X}|=|(1-\lambda) \vec{A}+\mu \vec{B}|=|1-\lambda+\mu \varepsilon| \cdot|\vec{A}|
$$

we get
$$
|1-\lambda+\mu \varepsilon|=1
$$

Then,
$$
1 \leqslant|1-\lambda|+|\mu \varepsilon|=1-\lambda+\mu
$$

Thus, $\mu>\lambda$, which means point $X$ will move on the extension of the line from $B$ to $O$. 

Next, consider the farthest distance of $X$ from $O$. By symmetry and continuity, the farthest distance of $X$ from $O$ should be achieved when $YB=ZB$. Therefore, $\lambda+\mu=1$ or $1-\lambda=\mu$. Let the side length of the regular $n$-sided polygon be $a$, and $OB$ intersects $XY$ at $R$. From
$$
\frac{Y R}{Y B}=\frac{a / 2}{\mu a}=\cos \frac{\pi}{n}
$$

we get $2 \mu=\frac{1}{\cos \frac{\pi}{n}}$. Thus, the farthest distance of $X$ from $O$ is
$$
\begin{array}{c}
(\mu-\lambda)|\vec{B}|=(2 \mu-1)|\vec{B}|=\left(\frac{1}{\cos \frac{\pi}{n}}-1\right) \frac{a}{2 \sin \frac{\pi}{n}}= \\
a\left(1-\cos \frac{\pi}{n}\right) / \sin \frac{2 \pi}{n}
\end{array}
$$

It is evident that the required trajectory is a "star shape" formed by drawing $n$ line segments from $O$ away from each vertex of the polygon, all of equal length, given by the above formula.