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Using a computer, I found the following result. Now, I'm looking for a solution to prove this result:

Let $A_{01}$, $A_{02}$, $A_{03}$, $A_{04}$, $A_{05}$, $A_{06}$ be six points in the plane, such that $A_{01}A_{02}A_{03}A_{04}A_{05}A_{06}$ is not a regular hexagon. Define $A_{{i+1}{j}}$ is midpoint of segment $A_{{i}{j}}A_{{i}{j+1}}$ where $j=\{1,2,\cdots,6\}$ here we take the modulo $6$ of the index $j$ and for $i=\{1,2,\cdots,n\}$ . Let $G_i$ be the centroid of the hexagon $A_{{i}{1}}A_{{i}{2}}A_{{i}{3}}A_{{i}{4}}A_{{i}{5}}A_{{i}{6}}$ for $i=\{1,2,\cdots,n\}$ then $G_1$, $G_2$, $G_3$...$G_{n}$ are collinear and $\lim_{n \to +\infty}\frac{G_{n+1}G_{n}}{G_nG_{n-1}}=\frac{1}{2}$

Example:

  • $A_{11}$ is midpoints of $A_{01}$$A_{02}$.
  • $A_{12}$ is midpoints of $A_{02}A_{03}$.
  • .................
  • $A_{16}$ is midpoints of $A_{06}A_{01}$.

and

  • $A_{21}$ is midpoints of $A_{11}$$A_{12}$;
  • $A_{22}$ is midpoints of $A_{12}A_{13}$
  • .................
  • $A_{26}$ is midpoints of $A_{16}A_{11}$

and

  • $A_{31}$ is midpoints of $A_{21}$$A_{22}$.
  • $A_{32}$ is midpoints of $A_{22}A_{23}$.
  • .................
  • $A_{36}$ is midpoints of $A_{26}A_{21}$.

and continuing.

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  • $\begingroup$ I think there is an interesting question here but the notation is very confusing. Are you treating the subscripts as two "digit" numbers where the first "digit" $i \in \{0,\dots,n\}$ is the iteration and the second $j \in \{1,\dots,6\}$ is the vertex? If so, each hexagon is obtained from the previous by cyclically connecting the midpoints of its sides. I suggest rephrasing the question in terms of midpoints to avoid the cumbersome and excessive notation. $\endgroup$ Commented May 7 at 13:55
  • $\begingroup$ Also, it's not clear whether you mean the (mass) centroid or the vertex centroid (see the Wikipedia article). In either case though, the limit in your question is not even defined in all cases, e.g., if the initial hexagon is regular, all the centroids coincide (and so $G_nG_{n-1} =0$ for all $n$). It would seem you need some additional non-degeneracy assumption. $\endgroup$ Commented May 7 at 14:00
  • $\begingroup$ Finally, it might be helpful (toward a proof or refutation) if you include some motivation for the conjecture. Why do you expect it to hold? It is an observed phenomenon in a computational experiment or do you have some mathematical reason to support the conjecture? $\endgroup$ Commented May 7 at 14:05
  • $\begingroup$ Aside from the degeneracy issue I mentioned above, I think you made a mistake in the limit assertion. (Maybe you meant the reciprocal?) As written, you're saying that the distance $G_{n+1}G_n$ between successive centroids essentially doubles at each step, but this cannot possibly be (if any of these distances as positive) as all the $G_i$ are contained in the convex hull of the initial hexagon. $\endgroup$ Commented May 7 at 14:12
  • $\begingroup$ @JackEdwardTisdell Thank you very much for several your excellent remarks above. I will improve my question. $\endgroup$ Commented May 8 at 1:50

1 Answer 1

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Based on your comment, it would seem that by centroid you mean the vertex centroid (as opposed to the mass centroid of the filled hexagon). But in this case, it's actually very easy to see that for any initial polygon whatsoever (including degenerate ones with coincident vertices and even non-planar closed polygonal curves), the sequence of vertex centroids obtained by taking the successive midpoints is in fact constant.

Generalizing your setup slightly, take $k$ initial points $A_{0,1},\dots,A_{0,k}$ in $\mathbb R^d$ (not necessarily distinct). Iteratively set $A_{n+1,j} = \frac{A_{n,j} + A_{n,j+1}}{2}$ to be the midpoint of the segment $A_{n,j}A_{n,j+1}$ (where $A_{n,k+1} = A_{n,1}$). Let $G_n = \frac{A_{n,1} + \cdots + A_{n,k}}{k}$ be the centroid of $A_{n,1},\dots,A_{n,k}$. (Yours is the case $k=6$ and $d=2$.)

Then $$ \begin{aligned} G_{n+1} &= \frac{A_{n+1,1}+A_{n+1,2}+\cdots+A_{n+1,k}}{k} \\&= \frac1k \Big( \frac{A_{n,1}+A_{n,2}}{2} + \frac{A_{n,2}+A_{n,3}}{2} + \cdots + \frac{A_{n,k}+A_{n,1}}{2}\Big) \\&= \frac1k( A_{n,1} + A_{n,2} + \cdots + A_{n,k} ) = G_n \end{aligned} $$ So by induction $G_n = G_0$ for all $n$, the centroid sequence is constant.

In particular, the $G_n$ are not merely colinear, they are all in fact the same point. The limit part of your question is not defined since the the numerator and denominator are always both zero. (As I mentioned in a comment, this $0/0$ issue was already obvious if you started with a regular hexagon but it is in fact generic, it doesn't matter what you start with.)

Note that, slightly more generally, the same argument works if you take a fixed convex or even affine combination at each step rather than the midpoints. That is, for any fixed sequence $\alpha_0,\alpha_1,\alpha_2,\dots$ of real numbers, set $A_{n+1,j} = \alpha_n A_{n,j} + (1-\alpha_n) A_{n,j+1}$. Exactly the same calculation as above shows that $G_n = G_0$.

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  • $\begingroup$ You were correct in all your comments and answers. I misunderstood how the software was using calculated the centroid. I have since looked into it again, and the software I use calculates the centroid according to the same formula you provided. I hope everyone will allow me to provide the correct definition of the centroid in this question $\endgroup$ Commented May 9 at 2:03
  • $\begingroup$ If my response answers your question, please "accept" my answer. Maybe the answer made you realize that the process you were interested in is not what you thought or not exactly what you asked about, this sometimes happens on this site! In that case, consider your problem carefully and post another question here if you feel it's appropriate. $\endgroup$ Commented May 9 at 6:42
  • $\begingroup$ Initially, I had a misunderstanding in identifying the centroid of the hexagon, so the question has been revised to accurately reflect what I intended to ask. Therefore, your answer is not yet accurate based on this revision. I hope you understand. Thank you sincerely $\endgroup$ Commented May 9 at 12:26
  • $\begingroup$ I am so sorry, I hope you delete your answer and I will delete my question. Then will post again because I think this is a difficult and good question. I will upvote you 10 times on your other questions and answers. @JackEdwardTisdell $\endgroup$ Commented Jun 6 at 8:46
  • $\begingroup$ This kind of spam upvoting is highly discouraged. I agree that the observed colinearity is non-trivial and interesting and there is nothing to stop you from posted a new (perhaps more carefully worded) question. But I will not delete a response which satisfactorily answered the original question (before editing). By the way, editing questions in substantial ways like this is also discouraged. Even if the original question turns out not to capture exactly was you had in mind, better to post a new question than to continually edit. $\endgroup$ Commented Jun 6 at 13:46

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