geometric center of points located on a regular polygon

Question

Suppose I have a regular 7 sided polygon. I have one point on each side of this polygon. The points cannot be on the vertices.

Can the geometric center of these 7 points also be the geometric center of another set of 7 points also located on the edges (vertices excluded) ?

Intuitively, I would say "no" because the number of edges is odd so points cannot be budged symmetrically, I would also think that the result shouldn't depend on the number of sides as long as it is odd

So the answer is "Yes". (see answers from jean Marie and Marty cohen)

Question 2 (asked by Marty Cohen and actually is the one I should have asked): what happens when the angles between consecutive lines from the center of the polygon to the 7 points are arbitrarily specified (the only restrictions being that they are positive and total 2π) ? Or if at least one of these angles is different from others ? Or in other words what happens when there is no symmetry between the points ? Does the "Yes" answer still hold ?

Thanks.

Answer 1

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 $2\pi$).

Answer 2

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.

Answer 3

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.

Answer 4

I am submitting this second answer since it is completely algebraic.

I show

(1) if the angular spacing between the lines is the same then the derived polygon (defined below) has the same center (this does not depend on the original polygon being regular)

and

(2) if the angular spacing linearly increases then the centers are not the same.

The points on a $n$-gon are $(z_k)_{k=1}^n$. Given $(r_k)_{k=1}^n$ with $0 < r_k < 1$ let the derived points $(w_k)_{k=1}^n$ be $w_k =r_kz_k+(1-r_k)z_{k-1} $ with $z_0 = z_n$.

Then

$\begin{array}\\ \sum_{k=1}^n w_k &=\sum_{k=1}^n (r_kz_k+(1-r_k)z_{k-1})\\ &=\sum_{k=1}^n r_kz_k+\sum_{k=1}^nz_{k-1}-\sum_{k=1}^nr_kz_{k-1}\\ &=\sum_{k=1}^nz_{k}+\sum_{k=1}^n r_kz_k-\sum_{k=1}^nr_{k+1}z_k\\ &=\sum_{k=1}^nz_{k}+\sum_{k=1}^n (r_k-r_{k+1})z_k\\ \end{array} $

Therefore, if $r_k=r_{k-1}$ for all $k$, the derived points have the same center.

Also, if the polygon is regular, so $z_k =e^{2\pi i k/n} $, then

$\begin{array}\\ \sum_{k=1}^n w_k &=\sum_{k=1}^n r_kz_k+\sum_{k=1}^nz_{k-1}-\sum_{k=1}^nr_kz_{k-1}\\ &=\sum_{k=1}^nz_{k}+\sum_{k=1}^n r_k(z_k-z_{k-1})\\ &=\sum_{k=1}^nz_{k}+\sum_{k=1}^n r_k(e^{2\pi i k/n}-e^{2\pi i(k-1)/n})\\ &=\sum_{k=1}^nz_{k}+\sum_{k=1}^n r_ke^{2\pi i (k-1)/n}(e^{2\pi i /n}-1)\\ &=\sum_{k=1}^nz_{k}+(e^{2\pi i /n}-1)\sum_{k=1}^n r_ke^{2\pi i(k-1)/n}\\ \end{array} $

so the derived center is the same as the original center if and only if $S(r) =\sum_{k=1}^n r_ke^{2\pi i(k-1)/n} =0 $.

For example, suppose $r_k = k/n$, so the spacing gradually increases. Then

$\begin{array}\\ S(r) &=\sum_{k=1}^n (k/n)e^{2\pi i(k-1)/n}\\ &=\dfrac1{n}\sum_{k=1}^n ke^{2\pi i(k-1)/n}\\ &=\dfrac1{n}\sum_{k=1}^n kx^{k-1} \qquad x=e^{2\pi i/n}\\ &=\dfrac1{n}\dfrac{n x^{n + 1} - (n + 1) x^n + 1}{(x - 1)^2}\\ &=\dfrac1{n}\dfrac{T(r)}{U(r)} \qquad T(r)=n x^{n + 1} - (n + 1) x^n + 1, U(r)=(x-1)^2\\ \\ T(r) &=n x^{n + 1} - (n + 1) x^n + 1\\ &=x^n(nx-(n+1)) + 1\\ &=e^{2\pi i n/n}(ne^{2\pi i/n}-(n+1)) + 1\\ &=(ne^{2\pi i/n}-(n+1)) + 1\\ &=n(e^{2\pi i/n}-1)\\ &\ne 0 \qquad\text{for } n > 1\\ U(r) &=(x-1)^2\\ &=(e^{2\pi i/n}-1)^2\\ \text{so}\\ S(r) &=\dfrac1{n}\dfrac{T(r)}{U(r)}\\ &=\dfrac1{n}\dfrac{n(e^{2\pi i/n}-1)}{(e^{2\pi i/n}-1)^2}\\ &=\dfrac1{e^{2\pi i/n}-1}\\ &\ne 0 \qquad\text{for } n > 1\\ \end{array} $