PlanetPhysics/Centre of Mass of Polygon

Let $A_{1}A_{2}{\ldots }A_{n}$ be an $n$-gon which is supposed to have a 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

${\begin{matrix}{\overrightarrow {OM}}={\frac {1}{n}}\sum _{i=1}^{n}{\overrightarrow {OA_{i}}}.\end{matrix}}$

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

${\begin{matrix}{\overrightarrow {AM}}={\frac {1}{3}}({\overrightarrow {AB}}+{\overrightarrow {AC}}).\end{matrix}}$

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

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