geometric center of points located on a regular polygon

If the lines from the center of the polygon to the points on the sides have the same angle between consecutive lines ($2\pi/n$), the resulting polygon will again be regular so its center will be the same as the original one.

I will leave it to others to work out what happens when the angles between consecutive lines are arbitrarily specified (the only restrictions being that they are positive and total $2\pi$).

The answer is "yes" (even if one excludes the elementary cases depicted in my first answer, and by a certain number of us).

Let $A_k$ ($k=0,\cdots 6$) be the vertices and $P_k$ the points on line segment $[A_k,A_{k+1}]$ (with a modulo 7 convention).

We have:

$$\underbrace{\frac17 \sum \vec{OP_k}}_{\text{c. of mass of points} \ P_k}=\underbrace{\frac17 \sum \vec{OA_k}}_{\text{c. of mass of points} \ A_k \ = \ O}+\frac17 \sum \vec{A_kP_k} \tag{1}$$

Let

$$v_k:=\vec{A_kP_k} \ \ \text{and} \ \ \vec{V}= \sum \vec{A_kP_k} \tag{2}$$

As a consequence of (1); the two centers of mass coincide if and only if

$$\vec{V}=0 \tag{3}$$

This is possible in a myriad of cases, as the following graphical representation shows. Indeed (3) is possible if and only, with the $v_k$s, one can built an heptagon of a special kind (irregular sidelengths and regular angles; i.e., all its internal angles are equal) with a "head-to-tail" loop representation (like in statics).

Such heptagons can be obtained graphically as the intersection of strips with variable width (or, if one prefers, of half-spaces) having polar angles $k \dfrac{2 \pi}{7}$. Slightly moving these strips in a orthogonal direction give different heptagons, i.e., different set of points $P_k$.

Now that we have understood the idea, a rigorous construction of any case is made possible by incremental construction. Here is how:

Let us assume that points $P_k$ are referenced by their abscissas $a_k$ defined by $\vec{A_kP_k}=a_k \vec{A_kA_{k+1}}.$ We will also consider that the sides have unit length.

• 0. Start with any case for values $a_k$ (for example all $a_k=1/2$ (midpoints)).
• 1. Make a choice of a certain index $k$ and a certain (small) value $d$.
• 2. change $a_k$ into $a_k+2d$ (the abscissa is either augmented of shortened according to the sign of $d$).
• 3. change $a_{k-1}$ into $a_{k-1}-d/\cos (2\pi/7)$.
• 4. change $a_{k+1}$ into $a_{k+1}-d/\cos (2\pi/7)$.

Then iterate the process 1)+2)+3)+4) (of course one has to check that abscissas always remain in the range $[0,1]$).

Remarks :

1. Angle $2\pi/7$ above is the so-called "external angle" of the heptagon.

2. A similar question could have been asked for any regular $n$-gon.

3. A question of the same type related to a general triangle (not equilateral in general): Coinciding centroids of two triangles.

4. As remarked by @dxiv, this center of mass preservation is not limited to the case where this center is $O$. It can be applied to any center of mass.

5. I end with a simulation result showing where centers of mass of points $P_k$ can be situated for random positions of these points on the sides (purple points) and for extreme cases where points $P_k$ are situated on some $A_k$s (red points).

Remark: A cousin issue, asked for a pentagon instead of a heptagon, is to be found here.

You can have the same the same center of mass.

Here is a counter example.

Let us take the notation $P_k, \ k=0\cdots 6$ for the vertices (modulo 7: for example $P_7=P_0$).

Conider the family of points $Q_k:=\frac23 P_k + \frac13 P_{k+1}$ (points situated at the limit of the first third on each side ; notations that can be understood with complex numbers for example) and the family of points $R_k:=\frac13 P_k + \frac23 P_{k+1}$ (points situated at the beginning of the last third on each side).

The $Q_k$ family has center of mass at the origin (proof: $\frac17 \sum (\frac23 P_k + \frac13 P_{k+1})=\frac17(\frac23 \sum P_k + \frac13 \sum P_{k+1}) = 0+0$).

Same computation for the other family $R_k$ : they share the same center of mass.