ATRACTOR

Consider an initial pentagon with vertices with abscissa \(x_0\), \(x_1\), \(x_2\), \(x_3\) and \(x_4\). For each \(x_r\) we write \[x_r\;=\;X+P_1\cos\frac{2r\pi}{5}+Q_1\sin\frac{2r\pi}{5}+ P_2\cos\frac{4r\pi}{5}+Q_2\sin\frac{4r\pi}{5}\] We have thus a system of \(5\) equations in \(5\) unknowns: \(X\), \(P_1\), \(Q_1\), \(P_2\) and \(Q_2\). 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 five equations, we get \[\sum_{r=0}^4 x_r\;=\;5X\] that is, \[X\;=\;\frac{1}{5}\sum_{r=0}^4 x_r\] and \(X\) represents the abscissa of the centroid of the \(5\) vertices. Note that, writing vector \((P_1,Q_1)\) in polar coordinates \((C_1 \cos\theta_1,C_1 \sin\theta_1)\) we have \[P_1\cos\frac{2r\pi}{5}+Q_1\sin\frac{2r\pi}{5}\;=\; C_1\cos\theta_1\cos\frac{2r\pi}{5} + C_1\sin\theta_1\sin\frac{2r\pi}{5}\;=\; C_1\cos\left(\frac{2r\pi}{5}-\theta_1\right)\] In a similar way, \[P_2\cos\frac{4r\pi}{5}+Q_2\sin\frac{4r\pi}{5}\;=\; C_2\cos\left(\frac{4r\pi}{5}-\theta_2\right)\] so that \[x_r\;=\;X+C_1\cos\left(\frac{2r\pi}{5}-\theta_1\right)+ C_2\cos\left(\frac{4r\pi}{5}-\theta_2\right).\] Then \[\begin{array}{ll}x'_r & =\;\frac{x_r+x_{r+1}}{2}\;=\\ & =\;\frac{1}{2}\left(X+C_1\cos\left(\frac{2r\pi}{5}-\theta_1\right)+ C_2\cos\left(\frac{4r\pi}{5}-\theta_2\right)+ X +\right.\\ &\;\;\;\;\;\left.C_1\cos\left(\frac{2(r+1)\pi}{5}-\theta_1\right)+ C_2\cos\left(\frac{4(r+1)\pi}{5}-\theta_2\right)\right)\;=\\ & =\;X+\frac{C_1}{2}\cos\left(\frac{2r\pi}{5}-\theta_1\right)+ \frac{C_1}{2}\cos\left(\frac{2r\pi}{5}-\theta_1+\frac{2\pi}{5}\right)+ \\&\;\;\;\;\; \frac{C_2}{2}\cos\left(\frac{4r\pi}{5}-\theta_2\right)+ \frac{C_2}{2}\cos\left(\frac{4r\pi}{5}-\theta_2+\frac{4\pi}{5}\right) \;=\\ & =\;X+C_1\cos\frac{\pi}{5} \cos\left(\frac{2r\pi}{5}-\theta_1+\frac{\pi}{5}\right) +C_2\cos\frac{2\pi}{5} \cos\left(\frac{4r\pi}{5}-\theta_2+\frac{2\pi}{5}\right) \end{array}\] Furthermore, \[\begin{array}{ll}x^{''}_r & =\;\frac{x'_r+x'_{r+1}}{2}\;=\\ & =\;X+C_1\cos^2\left(\frac{\pi}{5}\right) \cos\left(\frac{2r\pi}{5}-\theta_1+\frac{2\pi}{5}\right)+ C_2\cos^2\left(\frac{2\pi}{5}\right) \cos\left(\frac{4r\pi}{5}-\theta_2+\frac{4\pi}{5}\right) \end{array}\] and, more generally, \[\begin{array}{ll}x^{(k)}_r & =\;\frac{x^{(k-1)}_r+x^{(k-1)}_{r+1}}{2}\;=\\ & =\;X+C_1\cos^k\left(\frac{\pi}{5}\right) \cos\left(\frac{2r\pi}{5}-\theta_1+\frac{k\pi}{5}\right)+ C_2\cos^k\left(\frac{2\pi}{5}\right) \cos\left(\frac{4r\pi}{5}-\theta_2+\frac{2k\pi}{5}\right) \end{array}\] When \(k\) tends to infinite, the summands \(C_1\cos^k\left(\frac{\pi}{5}\right) \cos\left(\frac{2r\pi}{5}-\theta_1+\frac{k\pi}{5}\right)\) and \(C_2\cos^k\left(\frac{2\pi}{5}\right) \cos\left(\frac{4r\pi}{5}-\theta_2+\frac{2k\pi}{5}\right)\) converge both to zero. However, the latter converges faster than the former and we may forget it and take the approximation \[x^{(k)}_r\;\approx\;X+ C_1\cos^k\left(\frac{\pi}{5}\right) cos\left(\frac{2r\pi}{5}-\theta_1+\frac{k\pi}{5}\right)\] for large values of \(k\). Making \(C=C_1\cos^k\left(\frac{\pi}{5}\right)\) and \(\theta=\theta_1-\frac{k\pi}{5}\), we get \[x^{(k)}_r\;\approx\;X+C\cos\left(\frac{2r\pi}{5}-\theta\right)\] that is, \[x^{(k)}_r\;\approx\;X+P\cos\frac{2r\pi}{5}+Q\sin\frac{2r\pi}{5}\] where \(P=C\cos\theta\) and \(Q=C\sin\theta\). Similarly, \[y^{(k)}_r\;\approx\;Y+R\cos\frac{2r\pi}{5}+S\sin\frac{2r\pi}{5}\] and, assuming points in 3-dimensional space, \[z^{(k)}_r\;\approx\;Z+T\cos\frac{2r\pi}{5}+U\sin\frac{2r\pi}{5}\] 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 pentagon.

Let \(f\) be the linear transformation with matrix \(\left(\begin{array}{cc}P& Q\\R& S\\T& U\end{array}\right)\). Then \(P^{(k)}_r \approx (X,Y,Z)+ f\left(\cos\frac{2r\pi}{5},\sin\frac{2r\pi}{5}\right)\), and the points \(\left(\cos\frac{2r\pi}{3},\sin\frac{2r\pi}{3}\right)\) are the vertices of a regular pentagon 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 pentagon obtained by applying \(f\) to that regular pentagon, followed by a translation by vector \((X,Y,Z)\). Thus, while the regular pentagon is inscribed in a circle of unit radius centered at the origin, this pentagon is inscribed in an ellipse centered at \((X,Y,Z)\), and the pairs of parallel line segments in the regular pentagon (each side is parallel to a diagonal) stay parallel in this pentagon, as you may observe in the figure below.

Each side of the pentagon inscribed on the ellipse is parallel to one of its diagonals

Note also that \[x^{(k+2)}_r\;\approx\;X+C_1\cos^{k+2}\left(\frac{\pi}{5}\right) \cos\left(\frac{2r\pi}{5}-\theta_1+\frac{(k+2)\pi}{2}\right)\] \[x^{(k+2)}_r-X\;\approx\;C_1\cos^k\left(\frac{\pi}{5}\right) \cos^2\left(\frac{\pi}{5}\right) \cos\left(\frac{2r\pi}{5}-\theta_1+\frac{k\pi}{5}+ \frac{2\pi}{5}\right)\] \[x^{(k+2)}_r-X\;\approx\;\cos^2\left(\frac{\pi}{5}\right) C_1 \cos^k\left(\frac{\pi}{5}\right) \cos\left(\frac{2(r+1)\pi}{3}-\theta_1+\frac{k\pi}{5}\right)\] But, since \[x^{(k)}_{r+1}-X\;\approx\;C_1\cos^k\left(\frac{\pi}{5}\right) \cos\left(\frac{2(r+1)\pi}{3}-\theta_1+\frac{k\pi}{5}\right)\] then \[x^{(k+2)}_r-X\;\approx\;cos^2\left(\frac{\pi}{5}\right)(x^{(k)}_{r+1}-X)\] Thus, the points produced by the application twice of the bisection procedure to a pentagon are, approximately, the points obtained by an homothety with center at the centroid of the pentagon and ratio \(cos^2\left(\frac{\pi}{5}\right)\). The approximation is all the better the higher the value of \(k\). This explains the fact that the pentagons in the sequence appear alternately with the same shape, only with smaller size.