Other procedures
Let us see other methods of construction of new polygons, applied to a given \(n\)sided polygon with vertices with abscissa \(x_i\), \(i\in \{0,1,\ldots,n1\}\).

Trisection: if, instead of taking the middlepoints of each side of the polygon, we take the points that divide each side into two segments, the first with length double of the second one, we get a new \(n\)sided polygon whose vertices have abscissa \[x'_i=\frac{x_i+2x_{i+1}}{3}\] where \(x_n=x_0\).

More generally, if we join the points that divide each side of the given polygon in two parts, with lengths \(p\) and \(1p\) times the length of the side, with \(0 < p < 1\), we get a new \(n\)sided polygon whose vertices have abscissa \[x'_i= (1p) x_i+ p x_{i+1}\] where \(x_n=x_0\). View it in the following app.

Instead of taking the middlepoints of the sides of the given polygon, we may take the middlepoints of the line segments that join alternating vertices, that is, the points of abscissa \(x_{i1}\) and \(x_{i+1}\) (where \(x_{1}=x_{n1}\) and \(x_n=x_0\)). In this case, we produce a new \(n\)sided polygon whose vertices have abscissa \[x'_i=\frac{x_{i1}+x_{i+1}}{2}.\] Try it with the following app.

For even \(n=2m\), we may consider the middlepoints of the line segments that join opposite vertices. More generally, we may link the points that divide each one of those diagonals in two parts with lengths \(p\) and \(1p\) times the length of the diagonal, with \(0 < p < 1\). In this case, we get a new \(n\)sided polygon whose vertices have abscissa \[x'_i= (1p) x_i+ p x_{m+i}\] with \(x_{n+i}=x_i\), for every \(i\in\{0,1,\ldots,m1\}\). Observe it in the following app.
Instead of taking the arithmetic mean of the abscissa of two consecutive points, we may consider the mean of the abscissa of three (or more) consecutive points. In this case, we get a \(n\)sided polygon with vertices with abscissa \[x'_i=\frac{x_i+x_{i+1}+x_{i+2}}{3}\] where \(x_n=x_0\) and \(x_{n+1}=x_1\).
Note: the last case applied to quadrilaterals is very similar to bisection of triangles. If we consider \(x_m=\frac{x_0+x_1+x_2+x_3}{4}\), we have \[\begin{array}{ll} x'_0= & \frac{4}{3} x_m  \frac{1}{3} x_3\\ x'_1= & \frac{4}{3} x_m  \frac{1}{3} x_0\\ x'_2= & \frac{4}{3} x_m  \frac{1}{3} x_1\\ x'_3= & \frac{4}{3} x_m  \frac{1}{3} x_2 \end{array}\] that is, \[\begin{array}{ll} x'_0x_m =  \frac{1}{3} (x_3x_m)\\ x'_1x_m =  \frac{1}{3} (x_0x_m)\\ x'_2x_m =  \frac{1}{3} (x_1x_m)\\ x'_3x_m =  \frac{1}{3} (x_2x_m) \end{array}\] This means that the new points are obtained from the previous ones by an homothety of ratio \(1/3\) with center in a point \(G\) whose coordinates are given by the arithmetic mean of the corresponding coordinates of the four vertices of the given initial polygon. If these vertices are non complanar, they may be regarded as the vertices of a tetrahedron. Thereby, this process produces a sequence of embedded tetrahedrons in space, all similar to each other, with an increasingly smaller size, since the similarity ratio tends to zero.
Observe it in the following applet.