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According to Petr-Douglas-Neumann theorem (Wikipedia article) after all n-2 steps (n-number of sides/vertices in the polygon) are complete, the result is a regular polygon whose centroid coincides with the centroid of the original polygon.

Trying to check with GeoGebra the theorem for a quadrilateral, applying angle $\pi$ (k=2, midpoints of sides) and then angle $\frac{\pi}{2}$ ($k=1$, base angles $\frac{\pi}{4}$).

As the result I really get a square, but its centroid (i.e. centre) doesn't match the centroid of the original quadrilateral. To find the centroid of the original quadrilateral I use method suggested at another math.stackexchange thread.

BTW, when I connect the opposite midpoints of the original quadrilateral the intersection does coincide with the centre of constructed square. However, as far as I understand it, the point obtained by connecting opposite midpoints isn't in general a centroid.

At the attached image: $A_1 A_2 A_3 A_4$ — original quadrilateral, $B_1 B_2 B_3 B_4$ — after first step (applied angle $\pi$), $C_1 C_2 C_3 C_4$ — the result, $N$ — centroid of the original quadrilateral, $O$ — centroid of the resulting square.

GeoGebra file: https://drive.google.com/file/d/19BOvzUcFPmxl_d6zcKE6Bi-7yxmpItGC/view?usp=sharing

enter image description here

cyanide
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  • How are you calculating your centroid? In the context of PDN, "centroid" is simply the average of the vertices. – Blue Apr 26 '21 at 08:23
  • Indeed, there are actually two types of centroids: vertex centroid (the average of vertices' coordinates) and area centroid which is equivalent to centre of mass in physics. All the definitions of PDN theorem, I've come across, mention just "centroid", and "by default" this would mean area centroid. I looked though the paper by Jesse Douglas, it doesn't define the centroid either. Can you give a reference to a site that explicitly specifies that PDN means vertex centroid? – cyanide Apr 26 '21 at 09:21
  • "Can you give a reference to a site that explicitly specifies that PDN means vertex centroid?" ... You yourself have shown that the area-centroid doesn't match, so that can't be the proper interpretation. As PDN constructions as complex-number or matrix operations operate on vertices, it should be plausible that vertex-centroid is intended; indeed, in those contexts, the vertex-centroid result falls out fairly readily. ... I suppose you could simply take my word for it; my masters thesis was on generalizing a related result for graphs, and my advisor was (name drop) Branko Grunbaum. – Blue Apr 26 '21 at 09:47
  • Thanks. What you said really looks logical. Unfortunately, I can't mark that as an answer :) – cyanide Apr 26 '21 at 10:18
  • I'm intending to look-up a proper reference for you, and/or demonstrate the vertex-centroid result directly from complex/matrix methods, and put that into an answer. "Soon." – Blue Apr 26 '21 at 10:25

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In the context of Petr-Douglas-Neumann (PDN), "centroid" means "vertex-centroid" (ie, the average of the vertices). To see why the preservation of the vertex-centroid is "obvious" (and the least-interesting thing about PDN), let's consider a $n$-gon $p_0p_1p_2\cdots p_{n-1}$ with vertices $p_k$ in the complex plane.


Each step of the PDN process effectively applies a common operation to rotate-and-scale each side $p_kp_{k+1}$ about vertex $p_k$; we then shift focus to consider the polygon whose vertices are the images of the $p_{k+1}$. Each such image is given by $$p_k + t (p_{k+1}-p_k) = (1-t) p_{k}+ tp_{k+1} \tag1$$ for some complex $t := s e^{i\theta}$, where $s$ is the (real) scale factor and $\theta$ the (real) rotation angle. (The exact values of, or relations between, $s$ and $\theta$ don't matter for this discussion.) We can represent this effect across the entire polygon by using matrices. Defining $$P := \begin{bmatrix}p_0 & p_1 & p_2 & \cdots & p_{n-1}\end{bmatrix} \qquad\text{and}\qquad T_t := \begin{bmatrix} 1-t & 0 & 0 & \cdots & t \\ t & 1-t & 0 & \cdots & 0 \\ 0 & t & 1-t & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1-t \end{bmatrix} \tag2$$ we have that the entries of $PT_t$ are the vertices of the resulting polygon; likewise, the entries of $PT_t T_u$, $PT_t T_u T_v$, etc, are the vertices of polygons from further iterations of the process.

Now, the vertex-centroid of a polygon is the average of its vertices; ie, $$k := \frac1n(p_0+p_1+p_2+\cdots+p_{n-1}) = \frac1n P \,\mathbf{1}_n \tag3$$ where $\mathbf{1}_n$ is the "all-$1$s" column vector. But note: $T_t\,\mathbf{1}_n$ is just $\mathbf{1}_n$ again, so the vertex-centroid of the polygon obtained from applying the $T_t$ operation to $P$ is $$\frac1n(P\,T_t)\,\mathbf{1}_n = \frac1n P\,(T_t\,\mathbf{1}_n) = \frac1n P\,\mathbf{1}_n = k \tag4$$ That is to say: the vertex-centroid of the original polygon $P$ is preserved under the $T_t$ operation; and clearly, it's preserved under every $T$-operation applied thereafter. The thing never moves! $\square$


Of course, centroid-preservation is a worthwhile property to mention, but its significance pales in comparison to the fact that the PDN process ultimately results in a (standard, convex) regular $n$-gon from any arbitrary $n$-gon, thus generalizing Napoleon's theorem for triangles. Also interesting: (1) the order of the steps doesn't matter; and (2) if we replace a step with one based on that "final" $n$-gon, then the new "final" figure corresponds to the step we replaced! (I'm being a bit loosey-goosey with the description here. In any case, showing these aspects requires getting into the details of the various $s$- and $\theta$-values, so I'll leave that as an exercise to the reader.)

Blue
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