ATRACTOR

Given a triangle of vertices \(P_0\), \(P_1\) and \(P_2\), with abscissas \(x_0\), \(x_1\) and \(x_2\), we obtain, by bisection, a new triangle whose abscissas are given by \[x'_0\;=\;\frac{x_0+x_1}{2}\] \[x'_1\;=\;\frac{x_1+x_2}{2}\] \[x'_2\;=\;\frac{x_2+x_0}{2}\] Thereby, if we consider \(x_m=\frac{x_0+x_1+x_2}{3}\), we have \[x'_0\;=\;\frac{3}{2}x_m-\frac{1}{2}x_2\] \[x'_1\;=\;\frac{3}{2}x_m-\frac{1}{2}x_0\] \[x'_2\;=\;\frac{3}{2}x_m-\frac{1}{2}x_1\] that is, \[x'_0-x_m\;=\;-\frac{1}{2}(x_2-x_m)\] \[x'_1-x_m\;=\;-\frac{1}{2}(x_0-x_m)\] \[x'_2-x_m\;=\;-\frac{1}{2}(x_1-x_m)\] This means that the new points are obtained from the previous ones by means of a homothety of ratio \(-1/2\) and center at a point \(G\), called the barycenter, whose coordinates are given by the arithmetic mean of each of the respective coordinates of the three vertices of the initial triangle. Since this point belongs to the segments joining the midpoints of each side to the opposite vertices, to determine it, it suffices to consider the point of intersection of those three segments (in fact, two of those segments are enough).

It is easy to see why the new triangle is similar to the first triangle and moreover why it is obtained from the first one by a rotation of \(180\) degrees. This happens because each triangle is obtained from the previous one by a homothety of center in \(G\) and negative ratio. Any of the new triangles is similar to the previous one and therefore similar to the first triangle, and the similarity ratio is a power of \(1/2\). Moreover the triangles are getting smaller and smaller, since the similarity ratio tends to zero. However, apart from their size, all triangles look the same and are alternately at the same position of the first triangle or rotated by an angle of \(180\) degrees. Is this true for any polygons?