ATRACTOR

If we consider an hexagon with vertices with abscissa \(x_0\), \(x_1\), \(x_2\), \(x_3\), \(x_4\) and \(x_5\) we obtain a new hexagon whose vertices have abscissa \(x'_0\), \(x'_1\), \(x'_2\), \(x'_3\), \(x'_4\) and \(x'_5\), with \(x'_r=\frac{x_r+x_{r+1}}{2}\) for \(r\in\{0,1,2,3,4,5\}\) and \(x_6=x_0\).

The abscissa of the barycenter of the triangle with vertices \(x'_0\), \(x'_2\) and \(x'_4\) is given by \[\frac{x'_0+x'_2+x'_4}{3}=\frac{1}{3} (\frac{x_0+x_1}{2}+\frac{x_2+x_3}{2}+\frac{x_4+x_5}{2})= \frac{x_0+x_1+x_2+x_3+x_4+x_5}{6} \] while the abscissa of the barycenter of the triangle defined by vertices \(x'_1\), \(x'_3\) and \(x'_5\) is given by \[\frac{x'_1+x'_3+x'_5}{3}=\frac{1}{3} (\frac{x_1+x_2}{2}+\frac{x_3+x_4}{2}+\frac{x_5+x_0}{2})= \frac{x_0+x_1+x_2+x_3+x_4+x_5}{6} \] Likewise, the other coordinates also coincide. Thus, the barycenters of the two triangles are the same point. Its coordinates are given by the arithmetic mean of each one of the corresponding coordinates of the six vertices of the initial hexagon (this point, called gravity center or centroid, is the same for each one of the hexagons in the sequence). Moreover, the initial hexagon does not univocally determine the new hexagons, that is, the same sequence of hexagons can be produced from different initial hexagons. In fact, likewise in the case with quadrilaterals, there is an infinite number of initial hexagons that produces the same sequence of hexagons.

For each \(x_r\), \(r\in\{0,1,2,3,4,5\}\), we write \[x_r\;=\;X+P_1 \cos\frac{r\pi}{3} +Q_1 \sin\frac{r\pi}{3} +P_2 \cos\frac{2r\pi}{3} +Q_2 \sin\frac{2r\pi}{3} +P_3 \cos r\pi\] We have thus a system of \(6\) equations in \(6\) unknowns: \(X\), \(P_1\), \(Q_1\), \(P_2\), \(Q_2\) and \(P_3\). This is a system with a unique solution that can be computed from the information about abscissa. For example, to compute the value of \(X\) it suffices to see that, adding the six equations, we get \[\sum_{r=0}^5 x_r = 6X\] that is, \[X \;=\;\frac{1}{6} \sum_{r=0}^5 x_r\] and \(X\) represents the abscissa of the centroid of the \(6\) vertices. Note that, writing vector \((P_1,Q_1)\) in polar coordinates \((C_1 \cos\theta_1,C_1 \sin\theta_1)\), we have \[P_1\cos\frac{r\pi}{3}+Q_1\sin\frac{r\pi}{3}\;=\; C_1\cos\theta_1\cos\frac{r\pi}{3} + C_1\sin\theta_1\sin\frac{r\pi}{3}\;=\; C_1\cos\left(\frac{r\pi}{3}-\theta_1\right)\] Similarly, \[P_2\cos\frac{2r\pi}{3}+Q_2\sin\frac{2r\pi}{3}\;=\; C_2\cos\left(\frac{2r\pi}{3}-\theta_2\right)\] so that \[x_r\;=\;X+C_1\cos\left(\frac{r\pi}{3}-\theta_1\right)+ C_2\cos\left(\frac{2r\pi}{3}-\theta_2\right)+ P_3\cos r\pi\] Then \[\begin{array}{ll}x'_r & =\;\frac{x_r+x_{r+1}}{2}\;=\\ & =\;\frac{1}{2}\left(X+C_1\cos\left(\frac{r\pi}{3}-\theta_1\right)+ C_2\cos\left(\frac{2r\pi}{3}-\theta_2\right)+ P_3\cos r\pi + X +\right.\\ & \;\;\;\;\;\left.C_1\cos\left(\frac{(r+1)\pi}{3}-\theta_1\right)+ C_2\cos\left(\frac{2(r+1)\pi}{3}-\theta_2\right)+ P_3\cos((r+1)\pi)\right)\;=\\ & =\;X+\frac{C_1}{2}\cos\left(\frac{r\pi}{3}-\theta_1\right)+ \frac{C_1}{2}\cos\left(\frac{r\pi}{3}-\theta_1+\frac{\pi}{3}\right)+\\ & \;\;\;\;\;\frac{C_2}{2}\cos\left(\frac{2r\pi}{3}-\theta_2\right)+ \frac{C_2}{2}\cos\left(\frac{2r\pi}{3}-\theta_2+\frac{2\pi}{3}\right)+\\ & \;\;\;\;\; P_3\cos r\pi + P_3\cos(r\pi+\pi)\;=\\ & =\;X+C_1\cos\left(\frac{r\pi}{3}-\theta_1+\frac{\pi}{6}\right) \cos\frac{\pi}{6}+\\ & \;\;\;\;\; C_2\cos\left(\frac{2r\pi}{3}-\theta_2+\frac{\pi}{3}\right) \cos\frac{\pi}{3}+ 2P_3\cos\left(r\pi+\frac{\pi}{2}\right)\cos\frac{\pi}{2}\;=\\ & =\;X+\frac{\sqrt{3}}{2} C_1\cos\left(\frac{r\pi}{3}-\theta_1+\frac{\pi}{6}\right)+ \frac{1}{2}C_2\cos\left(\frac{2r\pi}{3}-\theta_2+\frac{\pi}{3}\right) \end{array}\] Note that coefficient \(P_3\) vanishes, and so its value has no effect in the new produced hexagon. Hence, there is an infinite number of initial hexagons that produce the same sequence of hexagons.

Furthermore, \[\begin{array}{ll}x^{''}_r\;= & =\;\frac{x'_r+x'_{r+1}}{2}\;=\\ & =\;X+\frac{3}{4} C_1\cos\left(\frac{r\pi}{3}-\theta_1+\frac{\pi}{3}\right)+ \frac{1}{4} C_2\cos\left(\frac{2r\pi}{3}-\theta_2+\frac{2\pi}{3}\right) \end{array}\] and, more generally, \[\begin{array}{ll}x^{(k)}_r & =\;\frac{x^{(k-1)}_r+x^{(k-1)}_{r+1}}{2}\;=\\ & =\;X+\left(\frac{\sqrt{3}}{2}\right)^k C_1\cos\left(\frac{r\pi}{3}-\theta_1+\frac{k\pi}{6}\right)+ \left(\frac{1}{2}\right)^k C_2\cos\left(\frac{2r\pi}{3}-\theta_2+\frac{k\pi}{3}\right) \end{array}\] When \(k\) tends to infinite, the summands \(\left(\frac{\sqrt{3}}{2}\right)^k C_1\cos\left(\frac{r\pi}{3}-\theta_1+\frac{k\pi}{6}\right)\) and \(\left(\frac{1}{2}\right)^k C_2\cos\left(\frac{2r\pi}{3}-\theta_2+\frac{k\pi}{3}\right)\) converge both to zero. However, the latter converges faster than the former and we may forget it and take the approximation \[x^{(k)}_r\;\approx\;X+\left(\frac{\sqrt{3}}{2}\right)^k C_1\cos\left(\frac{r\pi}{3}-\theta_1+\frac{k\pi}{6}\right)\] for large values of \(k\). Making \(C=\left(\frac{\sqrt{3}}{2}\right)^k C_1\) and \(\theta=\theta_1-\frac{k\pi}{6}\), \[x^{(k)}_r\;\approx\;X+C\cos\left(\frac{r\pi}{3}-\theta\right)\] that is, \[x^{(k)}_r\;\approx\;X+P\cos\frac{r\pi}{3}+Q\sin\frac{r\pi}{3}\] where \(P=C\cos \theta\) and \(Q=C\sin \theta\). Similarly, \[y^{(k)}_r\;\approx\;Y+R\cos\frac{r\pi}{3}+S\sin\frac{r\pi}{3}\] and, assuming points in 3-dimensional space, \[z^{(k)}_r\;\approx\;Z+T\cos\frac{r\pi}{3}+U\sin\frac{r\pi}{3}\] Hence the points \(P^{(k)}_r=(x^{(k)}_r,y^{(k)}_r,z^{(k)}_r)\) converge to the (possibly degenerate) plane defined by point \((X,Y,Z)\) and vectors \((P,R,T)\) and \((Q,S,U)\). This plane is independent of \(k\), and can be determined from the original hexagon.

Let \(f\) be the linear transformation with matrix \(\left(\begin{array}{cc}P& Q\\R& S\\T& U\end{array}\right)\). Then \(P^{(k)}_r \approx (X,Y,Z)+ f\left(\cos\frac{r\pi}{3},\sin\frac{r\pi}{3}\right)\), and the points \(\left(\cos\frac{r\pi}{3},\sin\frac{r\pi}{3}\right)\) are the vertices of a regular hexagon inscribed in a circle of unit radius centered at the origin. Thereby, the points \(P_r^{(k)}\) get closer and closer to the vertices of the hexagon obtained by applying \(f\) to that regular hexagon, followed by a translation by vector \((X,Y,Z)\). Thus, while the regular hexagon is inscribed in a circle of unit radius centered at the origin, this hexagon is inscribed in an ellipse centered at \((X,Y,Z)\), and the parallel sides in the regular hexagon stay parallel in this hexagon, as you may observe in the figure below:

Each side of the hexagon inscribed on the ellipse is parallel to the opposite side and to one of its diagonals

Note also that \[x^{(k+2)}_r\;\approx\;X+\left(\frac{\sqrt{3}}{2}\right)^{k+2} C_1\cos\left(\frac{r\pi}{3}-\theta_1+\frac{(k+2)\pi}{6}\right)\] \[x^{(k+2)}_r-X\;\approx\;\left(\frac{\sqrt{3}}{2}\right)^k \left(\frac{\sqrt{3}}{2}\right)^2 C_1\cos\left(\frac{r\pi}{3}-\theta_1+\frac{k\pi}{6}+ \frac{\pi}{3}\right)\] \[x^{(k+2)}_r-X\;\approx\;\frac{3}{4}\left(\frac{\sqrt{3}}{2}\right)^k C_1\cos\left(\frac{(r+1)\pi}{3}-\theta_1+\frac{k\pi}{6}\right)\] But, since \[x^{(k)}_{r+1}-X\;\approx\;\left(\frac{\sqrt{3}}{2}\right)^k C_1\cos\left(\frac{(r+1)\pi}{3}-\theta_1+\frac{k\pi}{6}\right)\] we have \[x^{(k+2)}_r-X\;\approx\;\frac{3}{4}(x^{(k)}_{r+1}-X)\] Hence, the points that we obtain by applying twice the bisection process to an hexagon are, approximately, the points obtained by an homothety with center at the centroid of the hexagon and ratio \(3/4\). The approximation is all the better the higher the value of \(k\). This explains the fact that the hexagons in the sequence appear alternately with the same shape, only with smaller size.