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 4 points that are the vertices of a quadrilateral, and this quadrilateral has three sides that are sides of the given regular $n$-sided polygon. Then the minimum value of $k$ is $\qquad$ .
$8 k=\left[\frac{3 n}{4}\right]+1$ Hint: 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 sides of 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 excluding $A_{4 k}\left(k=1,2, \cdots,\left[\frac{n}{4}\right]\right)$ and $A_{n}$ points (if $4 \mid n$, then $A_{n}=$ $A_{4}\left[\frac{n}{4}\right]$). Clearly, there are no consecutive vertices of the regular $n$-gon in set $A$, hence $k \geqslant\left[\frac{3 n}{4}\right]+1$.

Conversely, in any set of $\left[\frac{3 n}{4}\right]+1$ vertices of the regular $n$-gon, there must be 4 consecutive vertices. That is, after deleting $\left[\frac{n-1}{4}\right]$ points from the $n$ points, since $\frac{\left[\frac{3 n}{4}\right]+1}{\left[\frac{n-1}{4}\right]}>3$, there must be a pair of consecutive deleted points that include 4 consecutive vertices, so $k \leqslant$ $\left[\frac{3 n}{4}\right]+1$.