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centre of mass of polygon

\documentclass{article}
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\begin{document}
Let $A_1A_2{\ldots}A_n$ be an \PMlinkname{$n$-gon}{Polygon} which is supposed to have a \PMlinkescapetext{constant} surface-density in all of its points, $M$ the centre of mass of the polygon and $O$ the origin.  Then the position vector of $M$ with respect to $O$ is
\begin{align}
\overrightarrow{OM} = \frac{1}{n}\sum_{i=1}^n\overrightarrow{OA_i}.
\end{align}
We can of course take especially\, $O = A_1$,\, and thus
$$\overrightarrow{A_1M} = \frac{1}{n}\sum_{i=1}^n\overrightarrow{A_1A_i} =  
\frac{1}{n}\sum_{i=2}^n\overrightarrow{A_1A_i}.$$

In the special case of the triangle $ABC$ we have
\begin{align}
\overrightarrow{AM} = \frac{1}{3}(\overrightarrow{AB}+\overrightarrow{AC}).
\end{align}
The centre of mass of a triangle is the common point of its medians.\\

\textbf{Remark.}  An analogical result with (2) concerns also the \PMlinkescapetext{homogeneous} tetrahedron $ABCD$,
$$\overrightarrow{AM} = \frac{1}{4}(\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}),$$
and any $n$-dimensional simplex (cf. the \PMlinkname{midpoint}{Midpoint} of line segment:\, $\overrightarrow{AM} = \frac{1}{2}\overrightarrow{AB}$).

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