I am looking for a proof that:


if $A\_{11}A\_{12}...A\_{1n}$; $A\_{21}A\_{22}...A\_{2n}$; $\cdots$; $A\_{i1}A\_{i2}...A\_{in}$; $\cdots$; $A\_{m1}A\_{m2}...A\_{mn}$ are $m$ oriented regular polygons ($n$-gons), where $n=2k$, then$\DeclareMathOperator\Area{Area}$
$$
\begin{align\*}
& \Area(A\_{11}A\_{21}...A\_{m1})+\Area(A\_{1\;k+1}A\_{2\;k+1}...A\_{m\;k+1})\\ =\ & \Area(A\_{12}A\_{22}...A\_{m2})+\Area(A\_{1\;k+2}A\_{2\;k+2}...A\_{m\;k+2})\\=\ & \cdots\\ =\ & \Area(A\_{1i}A\_{2i}...A\_{mi})+\Area(A\_{1\;k+i}A\_{2\;k+i}...A\_{m\;k+i})\\ =\ & \cdots
\end{align\*}
$$


**Reference:**


* [Areas que suman lo mismo](http://www.xente.mundo-r.com/ilarrosa/GeoGebra/AreasIg_Npoligonos.html)
* A case $m=4$ and $n=4$, I posed in [here](https://math.stackexchange.com/questions/1447773/two-conjectures-of-four-squares) is near six year ago. But no have a proof.



Equal area of sum of pair opposite polygons conjecture
One fixes the counterclockwise orientation for polygons. Let $\omega\_1\omega\_2\cdots\omega\_m$ ($m\geq 3$) be a polygon, where $\omega\_i$'s are its vertices represented by complex numbers. It is standard to show that the area of the polygon is given by the imaginary part of the expression $$-\frac 1 2\sum\_{i=1}^m\omega\_i\overline{\omega\_{i+1}},\qquad (1)$$ where $\omega\_{m+1}=\omega\_1.$


Since the vertices of a regular $n$-gon are obtained from the $n$-th roots of unity by suitable translation, rotation and scaling, one may assume that the vertices of $A\_{i1}A\_{i2}\cdots A\_{in}$ are represented by the complex numbers $$A\_{ij}=\alpha\_i+\beta\_i\zeta^{j-1},1\leq j\leq n,1\leq i\leq m,$$ where $n=2k~(k\geq 2),$ and $\zeta=e^{2\pi i/n}$ is an $n$-th root of unity. Note that $\zeta^k=-1.$


By (1), the area of $A\_{11}A\_{21}\cdots A\_{m1}$ (resp., $A\_{1~k+1}A\_{2~k+1}\cdots A\_{m~k+1}$) is represented as the imaginary part of $$-\frac 1 2\sum\_{i=1}^m\left(\alpha\_i+\beta\_i\right)\left(\overline{\alpha\_{i+1}}+\overline{\beta\_{i+1}}\right)$$
$$\left({\rm resp.,} -\frac 1 2\sum\_{i=1}^m\left(\alpha\_i+\beta\_i\zeta^k\right)\left(\overline{\alpha\_{i+1}}+\overline{\beta\_{i+1}\zeta^k}\right)\right),$$ where $\alpha\_{m+1}=\alpha\_1$ and $\beta\_{m+1}=\beta\_1$. Their sum (before taking the imaginary part) equals
$$-\frac 1 2\sum\_{i=1}^m\left[\left(\alpha\_i+\beta\_i\right)\left(\overline{\alpha\_{i+1}}+\overline{\beta\_{i+1}}\right)+\left(\alpha\_i-\beta\_i\right)\left(\overline{\alpha\_{i+1}}-\overline{\beta\_{i+1}}\right)\right]$$
$$=-\sum\_{i=1}^m\left(\alpha\_i\overline{\alpha\_{i+1}}+\beta\_i\overline{\beta\_{i+1}}\right),\qquad (2)$$ where one uses that fact that $\zeta^k=-1.$


Now replacing $A\_{i1}$ by $A\_{ij}$ (resp. $A\_{i~k+1}$ by $A\_{i~j+k}$) for $1\leq i\leq m$ amounts to replacing $\beta\_i$ by $\beta\_i\zeta^{j-1}$ (resp. $\beta\_i\zeta^k=-\beta\_i$ by $\beta\_i\zeta^{j+k-1}=-\beta\_i\zeta^{j-1}$) for $2\leq j\leq n$. This has no effect on (2), since $$\left(\beta\_i\zeta^{j-1}\right)\left(\overline{\beta\_{i+1}\zeta^{j-1}}\right)=\beta\_i\overline{\beta\_{i+1}}, {\rm etc.}$$ The original assertion follows after taking the imaginary part of (2).


