PlanetPhysics/Centre of Mass of Polygon

Let $$A_1A_2{\ldots}A_n$$ 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 $$\overrightarrow{A_1M} = \frac{1}{n}\sum_{i=1}^n\overrightarrow{A_1A_i} = \frac{1}{n}\sum_{i=2}^n\overrightarrow{A_1A_i}.$$

In the special case of the triangle $$ABC$$ we have $$\begin{matrix} \overrightarrow{AM} = \frac{1}{3}(\overrightarrow{AB}+\overrightarrow{AC}). \end{matrix}$$ 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:\, $$\overrightarrow{AM} = \frac{1}{2}\overrightarrow{AB}$$).