8. In a given regular $n(n \geqslant 10)$-sided polygon, choose $k$ points from the $n$ vertices such that among these $k$ points, there exist four points that are the vertices of a quadrilateral, and this quadrilateral has: one side that is a side of the given regular $n$-sided polygon. Then the minimum value of $k$ is $\qquad$ .
8. $\left[\frac{3 n}{4}\right]+1$.

Label the $n$ vertices of this regular $n$-gon in sequence as $A_{1}, A_{2}, \cdots, A_{n}$. A necessary and sufficient condition for a quadrilateral to have three sides as the given regular $n$-gon is that its four vertices are four consecutive vertices of the regular $n$-gon.

Let the set $A$ be the set of points excluding $A_{4 k}\left(k=1,2, \cdots,\left[\frac{n}{4}\right]\right)$ (where $[x]$ denotes the greatest integer not exceeding the real number $x$) and $A_{n}$ (if $41 n$, then $A_{n}=A_{4}\left[\frac{n}{4}\right]$). Clearly, the set $A$ contains no four consecutive vertices of the regular $n$-gon, hence
$$
k \geqslant\left[\frac{3 \pi}{4}\right]+1 \text {. }
$$

Conversely, in any set of $\left[\frac{3 n}{4}\right]+1$ points of the regular $n$-gon, there must be four consecutive vertices, i.e., after deleting $\left[\frac{n-1}{4}\right]$ points from the $n$ points. And
$$
\frac{\left[\frac{3 n}{4}\right]+1}{\left[\frac{n-1}{4}\right]}>3,
$$

thus, among the deleted points, there must be a set of four consecutive vertices.
Therefore, $k \leqslant\left[\frac{3 n}{4}\right]+1$.