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Difference between revisions of "Spieker center"

(Created page with "The Spieker center is defined as the center of mass of the perimeter of the triangle. The Spieker center of a <math>\triangle ABC</math> is the center of gravity of a homogene...")
 
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The Spieker center is defined as the center of mass of the perimeter of the triangle. The Spieker center of a <math>\triangle ABC</math> is the center of gravity of a homogeneous wire frame in the shape of <math>\triangle ABC.</math> The Spieker center is a triangle center and it is listed as the point <math>X_{10}.</math>
 
The Spieker center is defined as the center of mass of the perimeter of the triangle. The Spieker center of a <math>\triangle ABC</math> is the center of gravity of a homogeneous wire frame in the shape of <math>\triangle ABC.</math> The Spieker center is a triangle center and it is listed as the point <math>X_{10}.</math>
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== Incenter of medial triangle==
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[[File:Physical proof.png|400px|right]]
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Prove that <math>X_{10}</math> is the incenter of the medial triangle <math>\triangle DEF</math> of a <math>\triangle ABC.</math>
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<i><b>Proof</b></i>
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Let's hang up the <math>\triangle ABC</math> in the middle of side <math>BC.</math> Side <math>BC</math> is balanced.
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Let's replace side <math>AC</math> with point <math>E</math> (the center of mass of <math>AC,</math> the midpoint <math>AC).</math> Denote <math>\rho</math> the linear density of a homogeneous wire frame.
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The mass of point <math>E</math> is equal to <math>\rho \cdot AC,</math> the shoulder of the gravity force is <math>EE' = ED \sin \alpha = \frac {AB \sin \alpha }{2}.</math>
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The moment of this force is <math>g EE' \rho \cdot AC = \frac {g \rho\cdot AC \cdot AB}{2} \sin \alpha.</math>
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Similarly the moment gravity force acting on AB is <math> \frac {g \rho\cdot AC \cdot AB}{2} \sin \beta.</math>
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Therefore, equilibrium condition is <math>\alpha = \beta</math> and the center of gravity of a homogeneous wire frame  <math>ABC</math> lies on each bisector of <math>\triangle DEF.</math>
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This point is the incenter of the medial triangle <math>\triangle DEF.</math>
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'''vladimir.shelomovskii@gmail.com, vvsss'''

Revision as of 11:30, 7 August 2023

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

Physical proof.png

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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