ATRACTOR

Let \(P\) and \(Q\) be two plane polygons. If \(P=(P_0,P_1,\ldots,P_{n-1})\) and \(Q=(Q_0,Q_1,\ldots,Q_{n-1})\), with \(P_i=(x_i,y_i)\) and \(Q_i=(z_i,w_i)\), produce the same polygon, this means that \(Ax=Az\) and \(Ay=Aw\), with \(x=(x_0,x_1,\ldots,x_{n-1})\), \(y=(y_0,y_1,\ldots,y_{n-1})\), \(w=(w_0,w_1,\ldots,w_{n-1})\) and \(z=(z_0,z_1,\ldots,z_{n-1})\). Hence, \(A(x-z)=A(y-w)=0\), that is, the vectors \(x-z\) and \(y-w\) are in the kernel of matrix \(A\).
In the case of bisection, we have \(a_0=a_1=\frac{1}{2}\) and \(a_j=0\) for \(j > 1\) and the kernel of \(A\) is generated by vector \[u_{(n/2)}=((-1)^j)_{j=0,1,\ldots,n-1}=(1,-1,1,-1,\ldots).\] Thus, \(Q_i=P_i+R_i\) where \[R_i=(z_i-x_i,w_i-y_i)=((-1)^i k,(-1)^i k')=(-1)^i (k,k')\] for some real numbers \(k\) and \(k'\).
The area of \(P\) is given by \[\frac{1}{2}\left|\sum_{i=0}^{n-1} P_i\times P_{i+1}\right|\] with \((a,b)\times (c,d)=\det\left( \begin{array}{cc}a & b\\c & d\\\end{array}\right)=a d - b c\) and the are of \(Q\) is given by \[\begin{array}{ll}\frac{1}{2}\left|\sum_{i=0}^{n-1} Q_i\times Q_{i+1}\right| & =\; \frac{1}{2}\left|\sum_{i=0}^{n-1} (P_i+R_i)\times (P_{i+1}+R_{i+1})\right|\;=\\ & =\; \frac{1}{2}\left|\sum_{i=0}^{n-1} P_i\times P_{i+1} +\sum_{i=0}^{n-1} P_i\times R_{i+1} + \sum_{i=0}^{n-1} R_i\times P_{i+1} + \sum_{i=0}^{n-1} R_i\times R_{i+1}\right| \end{array}\] Since \[\begin{array}{ll}\sum_{i=0}^{n-1} P_i\times R_{i+1} & =\; \sum_{i=0}^{n-1} (x_i,y_i)\times (-1)^{i+1}(k,k')\;=\\ & =\; \sum_{i=0}^{n-1} (x_{i+1},y_{i+1})\times (-1)^i(k,k')\;=\\ & =\; \sum_{i=0}^{n-1} P_{i+1}\times R_i\;=\\ & =\; -\sum_{i=0}^{n-1} R_i\times P_{i+1} \end{array}\] and \[\sum_{i=0}^{n-1} R_i\times R_{i+1}\;=\;\sum_{i=0}^{n-1} (-1)^i(k,k')\times (-1)^{i+1}(k,k')\;=\;0\] we have then \[\frac{1}{2}\left|\sum_{i=0}^{n-1} Q_i\times Q_{i+1}\right|\;=\; \frac{1}{2}\left|\sum_{i=0}^{n-1} P_i\times P_{i+1}\right| \] that is, the areas of polygons \(P\) and \(Q\) are the same.

Alternatively, we could also have observed that the areas of the polygons obtained by linking the vertices of even index are equal, as well as the areas of the polygons obtained by linking the vertices of odd index. In fact, since \(Q_i=P_i+(k,k')\) for every even \(i\), the polygon obtained by linking the vertices of \(Q\) of even index is the translate of the polygon obtained by linking the vertices of \(P\) of even index by a translation vector \((k,k')\), and thus they have the same area. Similarly, the polygon defined by the vertices of \(Q\) of odd index is the translate of the polygon defined by the vertices of \(P\) of even index by a translation vector \((-k,-k')\), so they also have the same area. Finally, as the area of any polygon can be calculated from the values of the areas of the polygon constructed by bisection and the polygons obtained by linking, respectively, the vertices of even index and those of odd index, and these values are the same for polygons \(P\) and \(Q\), then their areas are also equal.