Let us consider $\boldsymbol{b}$ be the position vector of the centre of mass of a regular $n$-polygon.
$(n-1)$ equal point masses are placed at $(n-1)$ vertices of the regular $n$-polygon, then for its centre of mass $r_{c m}$ when $m$ is placed at $n$th vertex.
$$
r_{C M}=\frac{(n-1) m b+m a}{(n-1) m+m}
$$
If mass $m$ is placed at $n$th remaining vertex or mass lies at centre then
$$
\begin{aligned}
&r_{C M}=0 \\
&\frac{(n-1) m b+m a}{(n-1) m+m}=0 \\
&(n-1) m b+m a=0 \\
&\vec{b}=-\frac{\vec{a}}{(n-1)}
\end{aligned}
$$
(-) sign shows that $C M$ lies other side from $n$th vertex geometrical centre of $n$-polygon i.e., $\vec{b}$ is opposite to the vector $\vec{a}$ from centre to $n$th vertex.