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Difference between revisions of "Spieker center"

(→‎Incenter of medial triangle)
(→‎Intersection of three cleavers)
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We use notation of previous proof. <math>DG</math> is the segment contains the Spieker center, <math>G \in AB, \angle EDG = \angle FDG, H = DG \cap AC.</math> WLOG, <math>AC > AB.</math>  
 
We use notation of previous proof. <math>DG</math> is the segment contains the Spieker center, <math>G \in AB, \angle EDG = \angle FDG, H = DG \cap AC.</math> WLOG, <math>AC > AB.</math>  
 
<cmath>DF||AC \implies \angle AHG = \angle FDG.</cmath>
 
<cmath>DF||AC \implies \angle AHG = \angle FDG.</cmath>
−
Similarly, <math> \angle AGH = \angle EDG =  \angle AHG \implies AH = AG \implies CH = AB + AH \implies DH</math> is cleaver.
+
Similarly, <math> \angle AGH = \angle EDG =  \angle AHG.</math>
 +
 
 +
So <math>AH = AG \implies CH = AB + AH \implies DH</math> is cleaver.
 +
 
 
Therefore, the three cleavers meet at the Spieker center.
 
Therefore, the three cleavers meet at the Spieker center.
  
 
'''vladimir.shelomovskii@gmail.com, vvsss'''
 
'''vladimir.shelomovskii@gmail.com, vvsss'''

Revision as of 12:17, 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 the Spieker center of triangle $\triangle ABC$ 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

Intersection of three cleavers

Cleaver.png

Prove that the Spieker center is located at the intersection of the three cleavers of triangle. A cleaver of a triangle is a line segment that bisects the perimeter of the triangle and has one endpoint at the midpoint of one of the three sides.

Proof

We use notation of previous proof. $DG$ is the segment contains the Spieker center, $G \in AB, \angle EDG = \angle FDG, H = DG \cap AC.$ WLOG, $AC > AB.$ \[DF||AC \implies \angle AHG = \angle FDG.\] Similarly, $\angle AGH = \angle EDG = \angle AHG.$

So $AH = AG \implies CH = AB + AH \implies DH$ is cleaver.

Therefore, the three cleavers meet at the Spieker center.

vladimir.shelomovskii@gmail.com, vvsss

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