# Show that this structure is an affinely regular polygon

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.

$$\def\para{\mathrel{/\!/}}\def\paren#1{\left(#1\right)}$$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.

• Thank you, can you give a little bit more information on the generalization of the structure itself? Dec 3, 2021 at 7:24
• 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? Dec 3, 2021 at 21:29
• That would be great! Dec 4, 2021 at 2:20
• @polygonlink1 Now with a construction on $\mathbb Z^2$.
– Ѕааԁ
Dec 4, 2021 at 12:41
• This is great, thank you so much! Dec 5, 2021 at 3:00