I posed a generalization of Theorem 3.2 In my paper

Conjecture: Let $P_1, P_2,....,P_{2n+1}$ and $O$ be $2n+2$ points in plane. Construct a chain $2n+1$ regular ${2n+1}$-gons $A_{1\;1}A_{1\;2}...A_{1\;2n+1}$; ....;$A_{2n+1\;1}A_{2n+1\;2}...A_{2n+1\;2n+1}$ with center $A_1, A_2...., A_{2n+1}$ such that $A_{1\;1}=O$, $A_{1\;2}=P_1$, $A_{2\;1}=A_{1\;3}$, $A_{2\;2}=P_{2}$, $A_{i+1\;1}=A_{i\;3}$, $A_{i+1\;2}=P_{i+1}$ for $i=\overline{1\;2n}$

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Continuing construct a chain $2n+1$ regular ${2n+1}$-gons $B_{11}B_{12}...B_{1\;2n+1}$; ....;$B_{2n+1\;1}B_{2n+1\;2}...B_{2n+1\;2n+1}$ with centers $B_1, B_2...., B_{2n+1}$, such that $B_{1\;1}=A_{2n+1\;3}$, $B_{1\;2}=P_1$, $B_{2\;1}=B_{1\;3}$, $B_{2\;2}=P_{2}$, $B_{i+1\;1}=B_{i\;3}$, $B_{i+1\;2}=P_{i+1}$ for $i=\overline{1\;2n}$.

Then $B_{2n+1\;3}=O$ and segments $A_1B_1=A_2B_2=...=A_{2n+1}B_{2n+1}$ and $\angle (A_iB_i, A_{i+1}B_{i+1})=\frac{(2n-1)\pi}{2n+1}$

Corollary: $B_{2n+1\;3}$ is fixed point when $P_1$, $P_2$, ....,$P_n$ be moved.

Question 1: Is the conjecture correct?

Question 2: Let $P_1$, $P_2$, ....,$P_n$ are fixed point in the plane, find position of $O$ such that $A_{2n+1\;3}=O$

See also:

PS: In the conjecture, all regular polygon is same direction.

  • $\begingroup$ Nice question ! $\endgroup$ – Alkan Jul 27 '20 at 10:43
  • $\begingroup$ Thanks you, I correction @FedorPetrov $\endgroup$ – Đào Thanh Oai Jul 27 '20 at 14:01

You may easily calculate everything in complex numbers. Denote $m=2n+1$, $w=e^{2\pi i/n}$, $Q_i=A_{i,3}=A_{i+1,2}$ for $i=i,2,\ldots$. We may suppose that $P_{m+i}=P_i$ for $i=1,\ldots,m$ and we have one sequence of $2m$ polygons. Then we have to prove $Q_{2m}=O$ (let $O=0$ be the origin) and that $C_k:=A_{k+m}-A_k$ satisfy $C_{k+1}=-w C_k$. This follows from $Q_k-P_k=(P_k-Q_{k-1})w$, where $Q_0=0$. Dividing by $w^k$ this gives for the sequence $R_k:=(-1)^kQ_k/w^k$ the recurrence $R_k-R_{k-1}=(-1)^kP_k(1+w)/w^k=:x_k$. We have $x_{m+k}+x_k=0$, thus $R_{2m}=x_1+\ldots+x_{2m}=0$. Also $A_k=(Q_k-P_kw)/(1-w)$, thus $$C_k=A_{k+m}-A_k=\frac{1}{1-w}(Q_{k+m}-Q_k)=(-1)^{k}\frac{w^k}{1-w}(R_k+R_{m+k})= (-1)^{k}\frac{w^k}{1-w}(x_1+\ldots+x_k+x_1+\ldots+x_{m+k})= (-1)^{k}\frac{w^k}{1-w}(x_1+\ldots+x_m),$$ the result follows.

As for question 2, it asks when $x_1+\ldots+x_m=0$. Since $O$ is variable, we replace $P_i$ to $P_i-O$ and get the equation $$ \sum_{k=1}^m (-1)^k (P_k-O)/w^k=0 \Leftrightarrow O=\frac{1+w}2\sum_{k=1}^m (-1)^{k-1}P_k w^{-k}. $$


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