Example 4.4.6 (IMO 27) Let $O$ be the center of a regular $n$-gon $(n \geqslant 5)$, and let two adjacent vertices be denoted as $A$ and $B$. Triangle $XYZ$ is congruent to $\triangle OAB$. Initially, let $\triangle XYZ$ overlap with $\triangle OAB$, then move $\triangle XYZ$ on the plane so that points $Y$ and $Z$ both move along the perimeter $A B C D \cdots$ of the polygon for one complete revolution, while point $X$ remains inside the polygon. Find the locus of point $X$.
Place $O$ at the origin of the complex plane.
Let $A, B, C$ be three consecutive vertices of the $n$-sided polygon, arranged counterclockwise. Suppose at some moment during the movement, $Y$ is on the line segment $AB$, and $Z$ is on the line segment $BC$. By the section formula, we have
$Y = (1 - \lambda)A + \lambda B, \quad 0 \leq \lambda \leq 1$,
$Z = (1 - \mu)B + \mu C, \quad 0 \leq \mu \leq 1$.
Let $t = e^{i \pi / n}$, then $B = tA, C = tB = t^2A$.
Clearly, we should have $(Y - X)t = Z - X$.
Solving the above equation and substituting the expressions for $Z$ and $Y$, we get
$$
X = \frac{Z - Yt}{1 - t} = \frac{1}{1 - t}(\mu - \lambda)(C - B),
$$

Substituting $C = tB$ into the equation, we obtain $X = -(\mu - \lambda)B$.

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, by $|OA| = |XY|$, we have
$$
|A| = |Y - X| = |(1 - \lambda)A + \mu B| = |(1 - \lambda) + \mu t||A|.
$$

This gives $|(1 - \lambda) + \mu t| = 1$. Since $1 \leq |1 - \lambda| + |\mu t| = 1 - \lambda + \mu$, it follows that $\mu \geq \lambda$.
Thus, point $X$ will lie on the extension of the line from $B$ to $O$.
When is $X$ farthest from $O$? By symmetry and continuity, this occurs when the segments $YB$ and $ZB$ are of equal length.
This implies $\lambda + \mu = 1$, or $1 - \lambda = \mu$. Therefore, $\mu |1 + t| = 1$.
By the definition of $t$ and the modulus of a complex number, we have $\mu \cdot |1 + \cos \frac{2\pi}{n} + i \sin \frac{2\pi}{n}| = \mu \cdot |2 \cos^2 \frac{\pi}{n} + 2i \sin \frac{\pi}{n} \cos \frac{\pi}{n}| = 1, 2 \mu \cdot |\cos \frac{\pi}{n}| \cdot | \cos \frac{\pi}{n} + i \sin \frac{\pi}{n}| = 1, 2 \mu = (\cos \frac{\pi}{n})^{-1}$.
Thus, the farthest distance is $(\mu - \lambda)|B| = (2\mu - 1)|B| = (1 - \cos \frac{\pi}{n})(\cos \frac{\pi}{n})^{-1}|B|$.
If we denote the side length of the regular $n$-sided polygon by $a$, we have $|B| = a(2 \sin \frac{\pi}{n})^{-1}$.
Therefore, the farthest distance from $X$ to $O$ is $a(1 - \cos \frac{\pi}{n})(\sin \frac{2\pi}{n})^{-1}$.
The conclusion is: The desired trajectory is a "star-shaped" figure centered at $O$, extending a line segment from $O$ in the direction away from each vertex of the polygon, with the length given by the above formula.