# Construct closed chain of $k$-gon around $n$ points-$n, k$ are odd primes number

Question 1: I am looking for a proof of the conjectures 1, 2, 3 as follows?

Question 2: In conjecture 3, in general case, I can not give a formula of $$X$$. But I think, If $$n, k$$ are odd primes number then $$X=\frac{2nk}{gcd(k-2,2k)gcd(n,k)}$$, I checked with some small case. Can You give a general formula of $$X$$?

Consider $$n, k \ge 3$$ be two integer numbers, given $$n$$ general points $$P_1$$, $$P_2$$,....,$$P_n$$ and $$O$$ is arbitrary point in the plane, let $$P_{n+i}=P_i$$ for $$i=1,\ldots,....$$. Construct a chain of $$m$$ regular $$k$$-gon:

• Construct first $$k$$-gon: $$A_{1\;1}A_{1\;2}....A_{1\;k}$$ such that $$A_{1\;1}=O$$; $$A_{1\;2}=P_1$$, the centroid of the first $$k$$-gon is $$A_1$$

• Second $$k$$-gon: $$A_{2\;1}A_{2\;2}....A_{2\;k}$$ such that $$A_{2\;1}=A_{1\;3}$$; $$A_{2\;2}=P_2$$ the centroid of the second $$k$$-gon is $$A_2$$

• $$.................................$$

• $$i$$ th $$k$$-gon: $$A_{i\;1}A_{i\;2}....A_{i\;k}$$ such that $$A_{i+1\;1}=A_{i\;3}$$; $$A_{i+1\;2}=P_{i+1}$$ the centroid of the $$i$$ th $$k$$-gon is $$A_i$$

• $$.................................$$

• $$m$$ th $$k$$-gon: $$A_{m\;1}A_{m\;2}....A_{m\;k}$$ such that $$A_{m\;1}=A_{m-1\;3}$$; $$A_{m\;2}=P_{m}$$ the centroid of the $$m$$ th $$k$$-gon is $$A_m$$

and all regular polygon is same direction.

Definition: The chain is closed if exist $$m$$ such that $$A_{m\;3}=A_{1\;1}=O$$. The chain is open if no exist $$m$$ such that $$A_{m\;3}=A_{1\;1}=O$$

Conjecture 1: If $$n=\frac{2k}{gcd(k-2,2k)}$$ then the chain is opened.

Conjecture 2: If $$n\ne\frac{2k}{gcd(k-2,2k)}$$ then the chain is closed.

Conjecture 3: If the chain is closed then $$m=n.X$$ and $$X$$ points $$A_i, A_{n+i}, A_{2n+i},...,A_{nx+i}$$ be form $$X$$-gon for $$i=1, 2,...,n$$ which the centroid of the $$X$$-gon is fixed when we moved $$O$$, these regular polygon equal.

UPDATE GEOGEBRA SOFTWARE APPLET

• @AlexRavsky I corrected – Đào Thanh Oai Aug 17 '20 at 4:43
• By geogebra software applet, I see that the chain will closed or open that is not depent on $P_j$ and $O$, this result depends on $n, k$ @AlexRavsky – Đào Thanh Oai Aug 17 '20 at 5:30

We can consider all given points as complex numbers, points of the complex plane. Then, as far as I understood, for all integers $$m\ge 1$$, $$1\le j\le k$$ we we have $$A_{m,j}=A_m+(P_m-A_m)\xi^{j-2},$$
where $$\xi=\exp\frac{2\pi i}{k}$$.

Since $$A_{m+1,1}= A_{m,3}$$, we have

$$A_{m+1}+(P_{m+1}-A_{m+1})\xi^{-1}=A_m+(P_m-A_m)\xi.$$

It follows

$$A_{m+1}=-A_m\xi+\frac{P_m\xi^2-P_{m+1}}{\xi-1}.$$

This formula suggests that the chain can be closed for a specific choice of $$O$$ and $$P_m$$’s, but I guess that you are looking for a stable general pattern. So let’s look when the sequence $$\{A_m\}$$ is periodic.

Putting $$B_m=A_m(-\xi)^{-m}$$, we obtain

$$B_{m+1}=B_m+\frac{P_m\xi^2-P_{m+1}}{\xi-1}(-\xi)^{-m-1}.$$

Since the sequence $$\{P_n\}$$ has a period $$n$$, we have

$$B_{m+2n}-B_{m+n}=(B_{m+n}-B_{m}) (-\xi)^{-n},$$

that is

$$A_{m+2n}-A_{m+n}(1+(-\xi)^{n})+A_m(-\xi)^{n}=0.$$

An equation $$\lambda^2-(1+(-\xi)^{n})\lambda +(-\xi)^{n}$$ has roots $$1$$ and $$(-\xi)^{n}$$. The following cases are possible.

1)) $$(-\xi)^{-n}=1$$. This holds iff ($$n$$ is even and $$k|n$$) or ($$n$$ is odd, $$k$$ is even and $$k|2n$$). The theory of recurrence relations implies that $$A_{r n+m}=c_1(m) + c_2(m)r$$ for each $$r$$ and some constants $$c_1(m)$$ and $$c_2(m)$$ depending on $$m$$. If all $$c_2(m)$$ are zeroes then the sequence $$\{A_m\}$$ has a period $$n$$ (or its divisor). Otherwise the sequence $$\{A_m\}$$ is not periodic. Thus the sequence $$\{A_m\}$$ is periodic iff for each $$m$$ we have $$A_m=A_{m+n}$$. This can happen iff the choice of $$P_m$$’s is specific. Namely,

1.1)) If $$n$$ is even and $$k|n$$ then

$$0=A_{m+n}(-\xi)^{-m-n}- A_{m}(-\xi)^{-m}=B_{m+n}-B_{m}=\frac{1+\xi }{1-\xi}\sum_{j=1}^n P_{m+j} (-\xi)^{-(m+j)},$$

that is, $$(-\xi)^{-1}$$ is a root of a polynomial $$P(x)=\sum_{j=1}^n P_{j} x^j$$.

1.2)) If $$n$$ is odd, $$k$$ is even, and $$k|2n$$ then

$$0=A_{m+2n}(-\xi)^{-m-2n}- A_{m}(-\xi)^{-m}=B_{m+2n}-B_{m}=\frac{1+\xi }{1-\xi}\sum_{j=1}^{2n} P_{m+j} (-\xi)^{-(m+j)},$$

that is, $$(-\xi)^{-1}$$ is a root of a polynomial $$(1+x^n)P(x)=\sum_{j=1}^{2n} P_{j} x^j$$

2)) $$(-\xi)^{-n}\ne 1$$. (This case holds, in particular, when both $$n$$ and $$k$$ are odd). The theory of recurrence relations implies that $$A_{r n+m}=c_1(m) + c_2(m)(-\xi)^{nr}$$ for each $$r$$ and some constants $$c_1(m)$$ and $$c_2(m)$$ depending on $$m$$. If all $$c_2(m)$$ are zeroes then the sequence $$\{A_m\}$$ has a period $$n$$ (or its divisor). Otherwise $$-\xi$$ is a primitive $$q$$-th root of unity, where $$q=\cases{k, \mbox{ if }k\equiv 0\pmod 4\\ k/2, \mbox{ if }k\equiv 2\pmod 4\\ 2k, \mbox{ if }k\equiv 1,3\pmod 4}.$$

Remark that $$q=\frac{2k}{\gcd(k-2,2k)}=\frac{2k}{\gcd(k-2,4)}$$. Thus $$(-\xi)^n$$ is a primitive $$\tfrac{q}{\gcd(q,n)}$$-th root of unity, and so the sequence $$\{A_m\}$$ has a period $$\tfrac{qn}{\gcd(q,n)}=\operatorname{lcm}(q,n)$$ (or its divisor). Moreover, for each $$m$$, points $$\{A_{r n+m}: 0\le r\le q-1\}$$ are vertices of a $$q$$-qon.

Finally, recall that for each $$m\ge 1$$, $$1\le j\le k$$ we have $$A_{m,j}=A_m+(P_m-A_m)\xi^{j-2}$$. It follows that if the sequence $$\{A_m\}$$ has a period $$p$$ then for each fixed $$j$$ a sequence $$\{A_{m,j}\}$$ has a period $$\operatorname{lcm}(p,n)$$ (or its divisor).

• Can You help me give your conclude? @AlexRavsky – Đào Thanh Oai Aug 14 '20 at 10:15
• Can you see my new update and click there? note that $O$ is blue color, $P_j$ are red color. If the chain is closed, then it is not depend on $P_i$ or $O$ it depends on $n, k$ – Đào Thanh Oai Aug 15 '20 at 6:14
• I update the construct Figure i.stack.imgur.com/JKk9r.png – Đào Thanh Oai Aug 15 '20 at 6:50
• I am very happy if You write the paper. You decide Co-author or only your auhthor. There are many result from this configuration. My English is not good. You can contact with me via oaidt.evnpsc@gmail.com – Đào Thanh Oai Aug 22 '20 at 5:07
• You can select some journal ijgeometry.com or geometry-math-journal.ro – Đào Thanh Oai Aug 22 '20 at 5:12