centre of mass of polygon

Let $A_1{A}_2\mathrm{\dots }{A}_n$ be an $n$-gon (http://planetmath.org/Polygon) which is supposed to have a 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

$\overrightarrow{OM}={\displaystyle \frac{1}{n}}{\displaystyle \sum _{i=1}^n}\overrightarrow{O{A}_i}.$(1)

We can of course take especially  $O=A_1$,  and thus

$$\overrightarrow{A_{1}M}=\frac{1}{n}\sum _{i=1}^n\overrightarrow{A_1{A}_i}=\frac{1}{n}\sum _{i=2}^n\overrightarrow{A_1{A}_i}.$$

In the special case of the triangle $ABC$ we have

$\overrightarrow{AM}={\displaystyle \frac{1}{3}}(\overrightarrow{AB}+\overrightarrow{AC}).$(2)

The centre of mass of a triangle is the common point of its medians.

Remark. An analogical result with (2) concerns also the tetrahedron $ABCD$,

$$\overrightarrow{AM}=\frac{1}{4}(\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}),$$

and any $n$-dimensional simplex (cf. the midpoint (http://planetmath.org/Midpoint) of line segment:  $\overrightarrow{AM}=\frac{1}{2}\overrightarrow{AB}$).