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Let $\{x_1,\cdots,x_n\}$ be a set of convex points labeled in a cyclic order. I am trying to show that the following structure is equivalent with an affinely regular polygon:

Fix $j$ for some $j\in \{1,\cdots,n\}$. Then we know the following to be true: $$\{x_j,x_{j-1}\}\parallel \{x_{j+1},x_{j-2}\},\quad \{x_{j},x_{j-2}\}\parallel \{x_{j+1},x_{j-3}\}$$ where indices are modulo $n$, and in general $\{x_j,x_{j-a}\}\parallel \{x_{j+1},x_{j-a-1}\}$ for each $a\in \mathbb{Z}_{+}$.

Here $\parallel$ denotes parallelity among the given lines. How would I go about showing that it is an affinely regular polygon?

I know that by Nizette we know that if for all $j\in \mathbb{Z}$ it holds that $\{x_j,x_{j+1}\}\parallel \{x_{j-1},x_{j+2}\}$ and $\{x_j,x_{j+2}\}\parallel \{x_{j-1},x_{j+3}\}$ then the structure is equivalent to an affinely regular polygon. However, the problem in my case is that $j$ is fixed to be some integer in $\{1,\cdots,n\}$. If there could be a way of showing that Nizette's argument is true based on this given information, then we would be done. However, I am not sure how I would go about this.

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$\def\para{\mathrel{/\!/}}\def\paren#1{\left(#1\right)}$Note: The false claim for $n = 4$ has been deleted.

For a systematic construction of counter-examples for $n \geqslant 4$, the crucial idea is to utilize concyclic polygons since concyclicity automatically ensures convexity, and parallelism are reduced to equality of arc angles. Below is an explicit counter-example for $n \geqslant 4$:$$ P_k = \paren{ \cos\paren{ \frac{k - 1}{n - 1} π }, \sin\paren{ \frac{k - 1}{n - 1} π } }\ (1 \leqslant k \leqslant n). $$ A figure can be shown by pasting and running the following code here:

int n = 8;  //Set the number of vertices here

unitsize(sqrt(n) * cm);
real pen_w = linewidth(defaultpen);

pair P[];
P.cyclic = true;
guide g = nullpath;

draw(unitcircle, dashed);

/* Draw the polygon */
for(int i = 0; i < n; ++i) {
    P.push(dir(i * 180 / (n - 1)));
    g = g -- P[i];
    label("$P_{" + string(i + 1) + "}$", P[i], P[i]);
}
g = g -- cycle;
draw(g);

/* Draw the parallel chords */
real N = min((int)sqrt(n) + 3, n - 3);
for(int i = 2; i <= N; ++i)
    draw(P[0] -- P[-i], gray((i - 1) / N));
for(int i = 2; i <= N + 1; ++i)
    draw(P[1] -- P[-i], gray((i - 2) / N));

/* Draw the unparallel chords */
pen strike_pen = blue + linewidth(log(n / 2.0 + 1) * pen_w);
draw(P[0] -- P[1], strike_pen);
draw(P[2] -- P[-1], strike_pen);

For $3 \leqslant k \leqslant n - 1$, since$$ \overparen{P_1 P_2} = \overparen{P_{k - 1} P_k} = \frac{π}{n - 1}, $$ then $P_1 P_k \para P_2 P_{k - 1}$, so the condition holds with $j = 1$. However, $P_1 P_2$ is not parallel to $P_3 P_n$ because$$ \overparen{P_1 P_n} = π ≠ \overparen{P_2 P_3} = \frac{π}{n - 1}, $$ so $P_1 \cdots P_n$ is not an affinely regular polygon.


Another construction on $\mathbb{Z}^2$:$$ P_k = (k, k(k - 1))\quad (0 \leqslant k \leqslant n - 1). $$ Since $\overrightarrow{P_{k - 1} P_k} × \overrightarrow{P_k P_{k + 1}} = (1, 2(k - 1)) × (1, 2k) = 2 > 0$ for $1 \leqslant k \leqslant n - 2$, and\begin{gather*} \small\overrightarrow{P_{n - 2} P_{n - 1}} × \overrightarrow{P_{n - 1} P_0} = (1, 2(n - 2)) × (-(n - 1), -(n - 1)(n - 2)) = (n - 1)(n - 2) > 0,\\ \small\overrightarrow{P_{n - 1} P_0} × \overrightarrow{P_0 P_1} = (-(n - 1), -(n - 1)(n - 2)) × (1, 0) = (n - 1)(n - 2) > 0, \end{gather*} then $P_0\cdots P_{n - 1}$ is convex. For $3 \leqslant k \leqslant n - 1$,$$ \overrightarrow{P_0 P_k} × \overrightarrow{P_1 P_{k - 1}} = (k, k(k - 1)) × (k - 2, (k - 1)(k - 2)) = 0, $$ thus $P_0 P_k \para P_1 P_{k - 1}$. Therefore the condition holds with $j = 0$. But$$ \overrightarrow{P_0 P_1} × \overrightarrow{P_2 P_{n - 1}} = (1, 0) × (2, (n - 1)(n - 2) - 2) = n(n - 3) ≠ 0, $$ i.e. $P_0 P_1$ is not parallel to $P_2 P_{n - 1}$, so $P_0 \cdots P_{n - 1}$ is not an affinely regular polygon.

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  • $\begingroup$ Thank you, can you give a little bit more information on the generalization of the structure itself? $\endgroup$ Dec 3, 2021 at 7:24
  • $\begingroup$ Thank you the correction. I was wondering if there exists other constructions or would this be the only one? There would not exist a construction corresponding to the given rule in $\mathbb{Q}^2$, for example? $\endgroup$ Dec 3, 2021 at 21:29
  • $\begingroup$ That would be great! $\endgroup$ Dec 4, 2021 at 2:20
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    $\begingroup$ @polygonlink1 Now with a construction on $\mathbb Z^2$. $\endgroup$
    – Ѕааԁ
    Dec 4, 2021 at 12:41
  • $\begingroup$ This is great, thank you so much! $\endgroup$ Dec 5, 2021 at 3:00

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