G53 (27-4, Iceland) For a regular $n(n \geqslant 5)$-sided polygon centered at $O$, two adjacent vertices are denoted as $A$ and $B . \triangle X Y Z \cong \triangle O A B$. Initially, let $\triangle X Y Z$ coincide with $\triangle O A B$, then move $\triangle X Y Z$ on the plane so that points $Y, Z$ both move along the perimeter of the polygon while point $X$ moves inside the polygon. Find the trajectory of point $X$.
Let the radius of the circumscribed circle of a regular $n$-sided polygon be $R$. As shown in Figure 4-55, by the symmetry of the regular polygon, we only need to consider the movement of $\triangle XYZ$ from the position of $\triangle OAB$ to the position of $\triangle OBC$ (where $A, B, C$ are three consecutive vertices of the regular $n$-sided polygon).
It is easy to see that,
$$
\begin{aligned}
\angle Y X Z & =\angle A O B \\
& =180^{\circ}-2 \angle O B A \\
& =180^{\circ}-\angle A B C .
\end{aligned}
$$

Therefore, $X, Y, B, Z$ are concyclic, so $\angle X B Y=\angle X Z Y=\angle O B A$, which means $X, O$, and $B$ are collinear.
By the Law of Sines,
$$
\begin{aligned}
B X= & \frac{Y Z}{\sin \angle A B C} \cdot \sin \angle X Y B \\
& \leqslant \frac{A B}{\sin \angle A O B}=\frac{2 R \sin \frac{180^{\circ}}{n}}{\sin \frac{360^{\circ}}{n}} \\
& =R \sec \frac{180^{\circ}}{n} .
\end{aligned}
$$

Equality holds only when $\angle X Y B=\angle X Z B=90^{\circ}$, in which case $X B$ exactly bisects $Y Z$ perpendicularly.

By continuity and symmetry, the required trajectory is a "star" shape composed of $n$ rays centered at $O$, opposite to the lines connecting $O$ with each vertex, with a length of $R\left(\sec \frac{180^{\circ}}{n}-1\right)$.