PlanetPhysics/Centre of Mass of Polygon

Let $A_1{A}_2\dots A_n$ be an $n$-gon which is supposed to have a constant surface-density in all of its points, $M$ the [[../CenterOfGravity/|centre of mass]] of the polygon and $O$ the origin. Then the [[../PositionVector/|position vector]] of $M$ with respect to $O$ is

$\begin{array}{c}\overrightarrow{OM}=\frac{1}{n}\sum _{i=1}^n\overrightarrow{O{A}_i}.\end{array}$

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

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

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:\, $\overrightarrow{AM}=\frac{1}{2}\overrightarrow{AB}$).