Equal area of sum of pair opposite polygons conjecture 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:

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*Areas que suman lo mismo


*A case $m=4$ and $n=4$, I posed in here is near six year ago. But no have a proof.
 A: A comment (well, a few observations), not an answer but too long for that.

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*Given vector spaces $V,W$ ,a bilinear mapping from $V\times W$ into the reals is one which is linear in each variable separately. Crucial fact: such an operator is uniquely determined by its vales on elements $(x,y)$ where $x$ and $y$ range over bases (yes, I know that this is undergraduate maths but the OP insists on details).


*If $X$ and $Y$ are planar vectors and $L(X,Y)$ is the signed area of the square $ABCD$ erected on $A=X$ and $B=Y$, then this is bilinear in $X$ and $Y$ (easy to prove directly in coord8nates using the area formula or by a simple geometrical argument— not my fault—the OP insists on details).


*Using these facts, then the case of squares can be reduced to that of determining when two bilinear mappings coincide.  This can easily be done with a simple Mathematica programme or by using symmetry and the canonical basis for $2$-space.


*This method can be carried over  to the more general case of polygons  and to gazillions of natural generalisations but I  am too terrified of the OP‘s potential comments to even think about posting them here.
Added as an edit—the square case in more detail:
We assume that $A_¡$ is represented by the vector $X_¡$, $B_i$ by $Y_i$—then $C_i$ and $D_i$ are $Y_i+DY_i-DX_i$ and $X_i+DY_i-DX_i$ ($D$ denotes rotation through a right angle).
Now the area of quadrilateral $A_1A_2A_3A_4$ is $$ \frac 12 (X_1\wedge X_2+X_2\wedge X_3 +X_3\wedge X_4+X_4\wedge X_1)$$  with corresponding formulae for the other three quadrilaterals.  One can then plug these expressions into the desired equations and complete the proof by routine, if tedious, computations.  As I indicated previously, one can, if desired,  take advantage of the fact that the expressions are bilinear forms on the corresponding vector spaces.
The general case becomes a battle with indices.  If $D$ now denotes rotation through the exterior angle of the regular polygon and $X_i$,$Y_i$ are the vectors corresponding to $A_{i1}$ and $A_{i1}A_{i2}$, then $$ A_{ik}=X_{i}+Y_{i}+DY_i+\dots +D^{k-2}Y_i. $$
We now use the formula for the area of the polygon with vertices $Z_1,\dots,Z_p$,
$$  \frac 12 (Z_1\wedge Z_2+ \dots + Z_p\wedge Z_1),$$ substitute the appropriate expressions and collect terms.
This is just a sketch, I know, but hope that it will be useful to the OP.
A: 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).
