Empty Triangle!

The centre of mass $G$ of the triangular frame $ABC$ , with line density $\rho$ , is equal to the centre of mass of three point masses $\rho a$ , $\rho b$ , $\rho c$ at the points $D,E,F$ respectively, where $D,E,F$ are the midpoints of the lines $BC,AC,AB$ respectively (here we are simply replacing each side of the triangle by a point particle of the same mass located at the midpoint of each side). $\;\;$ Consider the centre of mass $X$ of the point masses at $E$ and $F$ . This point lies on the line segment $EF$ and the ratio $EX\,:\, XF$ is equal to $c\,:\, b$ . Since $ED = \tfrac12c$ and $FD = \tfrac12b$ , this is the same as the ratio $ED\,:\, DF$ . By the Angle Bisector Theorem, we deduce that $X$ lies on the angle bisector of $\angle EDF$ . Then $G$ is the centre of mass of the system consisting of a particle of mass $\rho a$ at $D$ and a particle of mass $\rho(b+c)$ at $X$ . Thus $G$ lies on the line segment $DX$ , and hence lies on the angle bisector of $\angle EDF$ . Similar arguments tell us that $G$ lies on the angle bisectors of $\angle DEF$ and $\angle DFE$ as well, and hence that $G$ is the incentre of $DEF$ .

The centre of mass of a triangular frame $ABC$ is therefore the incentre of the medial triangle. This point is known as the Spieker centre for the triangle $ABC$ .