Spieker center

The Spieker center is defined as the center of mass of the perimeter of the triangle. The Spieker center of a $\triangle ABC$ is the center of gravity of a homogeneous wire frame in the shape of $\triangle ABC.$ The Spieker center is a triangle center and it is listed as the point $X_{10}.$

Incenter of medial triangle

Prove that $X_{10}$ is the incenter of the medial triangle $\triangle DEF$ of a $\triangle ABC.$

Proof

Let's hang up the $\triangle ABC$ in the middle of side $BC.$ Side $BC$ is balanced.

Let's replace side $AC$ with point $E$ (the center of mass of $AC,$ the midpoint $AC).$ Denote $\rho$ the linear density of a homogeneous wire frame.

The mass of point $E$ is equal to $\rho \cdot AC,$ the shoulder of the gravity force is $EE' = ED \sin \alpha = \frac {AB \sin \alpha }{2}.$

The moment of this force is $g EE' \rho \cdot AC = \frac {g \rho\cdot AC \cdot AB}{2} \sin \alpha.$

Similarly the moment gravity force acting on AB is $\frac {g \rho\cdot AC \cdot AB}{2} \sin \beta.$

Therefore, equilibrium condition is $\alpha = \beta$ and the center of gravity of a homogeneous wire frame $ABC$ lies on each bisector of $\triangle DEF.$

This point is the incenter of the medial triangle $\triangle DEF.$

vladimir.shelomovskii@gmail.com, vvsss

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Spieker center

Incenter of medial triangle