Generalization of some plane geometry theorems

Question

Conjecture: Let $A_1, A_2,\dotsc,A_n$; $B_1, B_2,\dotsc,B_n$ and $C_1, C_2,\dotsc,C_n$ be $3n$ points in the plane such that $\angle{\overrightarrow{A_iB_i}, \overrightarrow{A_{i+1}B_{i+1}}}=\frac{2\pi}{n}$ and $|A_iB_i|=|A_{i+1}B_{i+1}|$ for $i=\overline{1,n}$ and taking subscripts modul $n$. Let $2n$ points $D_1, D_2,\dotsc,D_n$; $E_1, E_2,\dotsc,E_n$ in the plane and $\ell$ is a real number such that $\overrightarrow{C_iD_i}=\ell\overrightarrow{C_iA_i}$ and $\overrightarrow{C_iE_i}=\ell\overrightarrow{C_iB_i}$ for $i=\overline{1,n}$ then $D_1D_2\dots D_n$ is a regular $n$-gon $\Leftrightarrow$ $E_1E_2\dots E_n$ is a regular $n$-gon (in this case these two regulars $n$ gons have the same centroid).

This result is generalization of some results:

• Fermat-Dao-Nhi equilateral triangle in preamble before points X(33602).

• Napoleon theorem.

• Thébault's problem I.

Example: Let $A_1, A_2,\dotsc,A_5$; $B_1, B_2,\dotsc,B_5$ be $10$ points such that $\angle{\overrightarrow{A_iB_i}, \overrightarrow{A_{i+1}B_{i+1}}}=72^\circ$ and $|A_iB_i|=|A_{i+1}B_{i+1}|$ for $i=\overline{1,5}$ and taking subscripts modul $5$. Let $D_1D_2\dots D_3$ be a regular pentagon in the plane. Let $C_i$ be the reflection of $A_i$ in $D_i$ and $E_i$ be the midpoint of $B_iC_i$ then $E_1E_2\dotsc E_5$ is a regular pentagon (in this example we let $n=5$, $\ell=\frac{1}{2}$).

Question: Using my computer I checked the conjecture is true for $n=3,4,5$. Does the conjecture correct?

Answer 1

Use complex coordinates (with uppercase letters replaced to corresponding lowercase) such that $b_j-a_j=w^j$, $w=e^{2\pi i/n}$. Then $c_j-d_j=\ell(c_j-a_j)$, so $c_j(1-\ell)=d_j-\ell a_j$, analogously $c_j(1-\ell)=e_j-\ell b_j$, thus $d_j-\ell a_j=e_j-\ell b_j$, $e_j=d_j+\ell(b_j-a_j)$, $e_j=d_j+\ell w^j$. $D_1\ldots D_n$ is regular polygon (properly directed) iff $d_j=\alpha w^j+\beta$ for fixed complex $\alpha,\beta$. Thus, we see that $e_j$ satisfy the same rule with $\alpha+\ell$ instead of $\alpha$.