18. For a regular $n$-sided polygon $(n \geqslant 5)$ centered at point $O$, let two adjacent vertices be $A$ and $B$. $\triangle XYZ$ is congruent to $\triangle OAB$. Initially, $\triangle XYZ$ coincides with $\triangle OAB$. Then, $\triangle XYZ$ is moved on the plane such that points $Y$ and $Z$ both travel along the perimeter of the polygon for one complete round, while point $X$ remains inside the polygon. Find the trajectory of point $X$.
18. Since $\angle Y X Z+\angle Y B Z=\angle A O B+\angle A B C=\frac{n-2}{n} \pi+\frac{2}{n} \pi=\pi$, then $X, Y, B, Z$ are concyclic, thus $\angle X B Y=\angle X Z Y=\angle O B Y$, which means $X$ lies on the extension of $B O$.
Also, $S_{\triangle X Y B}=\frac{1}{2} B X \cdot B Y \sin \angle X Y B=\frac{1}{2} X Y \cdot B Y \sin \angle X Y B$, hence
$$
B X=\frac{X Y \sin \angle X Y B}{\sin \angle Y B X}=\frac{X Y \sin \angle X Y B}{\sin \frac{(n-2) \pi}{2 n}} .
$$

The range of $\angle X Y B$ is $\frac{(n-2) \pi}{2 n} \leqslant \angle X Y B \leqslant \pi-\frac{(n-2) \pi}{2 n}$.
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 $n$-gon. 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 $B O$ twice. Thus, the locus of point $X$ is a "star-shaped" figure composed of line segments of length $d$ extending from the center of the regular $n$-gon away from each vertex.