A naiv geometric question.

We consider shapes of triangles and plane quadrilaterals (i.e. these objects up to translation, rotation and scaling).

The shape of triangles ($T$) is 2-dimensional (two angles resp. 3 points with 6 coordinates minus 2 translationals degree of freedoms minus 1 rotational degree of freedom minus 1 scaling degree of freedom).

The shape of plane quadrilaterals ($Q$) is 4-dimensional (one unit side to fix the scale and two adjacent sides defined by two angles and two side lenghts resp. 4 points with 8 coordinates minus 2 translationals degree of freedoms minus 1 rotational degree of freedom minus 1 scaling degree of freedom).

For a polygon we can define three natural centers of mass: a) ACM, area center of mass, b) ECM, edges center of mass (edges carry a homogenous mass density per length unit), c) VCM, vertices center of mass (equal masses at vertices).

For a triangle $t \in T$ we have always $\mathrm{ACM}(t)=\mathrm{VCM}(t)$. Regarding ECM: Assuming the two 2-dimensional graphs $\{(t, \mathrm{ACM}(t)), t \in T\} \subset T \times \mathbb{R}^2$ and $\{(t, \mathrm{ECM}(t)), t \in T\} \subset T \times \mathbb{R}^2$ lie in a general position and thus intersect transversally, we expect the space of triangle shapes with ACM=ECM to be 2+2-4=0-dimensional. And in fact, the only triangle shape with ACM=ECM is the equilateral one.

For plane quadrilaterals we would assume the two 4-dimensional graphs $\{(q, \mathrm{ACM}(q)), q \in Q\} \subset Q \times \mathbb{R}^2$ and $\{(q, \mathrm{ECM}(q)), q \in Q\} \subset Q \times \mathbb{R}^2$ also to lie in a general position and thus intersect transversally. And thus we expect the dimension of plane quadrilateral shapes with ACM=ECM to be 4+4-6=2-dimensional. And the same for the other two pairs (ECM=VCM and ACM=VCM). Is this correct?

Intersecting the 2-dimensional surface ACM=ECM with the 2-dimensional surface ACM=VCM we would expect a 0-dimensional space.

However the space of all planar quadrilaterals with ACM=ECM=VCM seems to be 2-dimensional, namely all paralellogramms (you can choose one angle and ration of side lengths; they have a $180^{\circ}$-symmetry and so ACM=ECM=VCM.

Why does dimension counting not work (which subspaces are not in general position?) or where am I mistaken?

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    $\begingroup$ It would make more sense to me to consider the two equations $\operatorname{ACM}(q)=\operatorname{ECM}(q)$ and $\operatorname{ACM}(q)=\operatorname{VCM}(q)$ as defining two hypersurfaces in $Q$, and hence intersecting in a $2$-dimensional subspace. It seems that the two graphs do not intersect transversally, but I don't have a proof. $\endgroup$ – Servaes Mar 16 '19 at 14:54
  • $\begingroup$ Very interesting question. There is a possible connection with the area of mathematics called "Geometric probability" or "Stochastic geometry" with the emblematic book by Santalo ? Just a remark : I wouldn't say "The shape of plane quadrilaterals (Q) is 4-dimensional" but "The set of shapes of plane ...". $\endgroup$ – Jean Marie Mar 16 '19 at 20:40
  • $\begingroup$ One question : what about crossed quadrilaterals (non convex shape) ? Among them, a "bow-tie" shape also has ACM=ECM=VCM... $\endgroup$ – Jean Marie Mar 18 '19 at 22:02

This is not a direct answer to the issue. Just information about different ways to consider the general framework.

1) You describe a quadrilateral shape as an equivalence class of quadrilaterals up to general similitudes, i.e., translations, homotheties and scaling. The presentation given in the "Hamburg school" directed in the '1930s by a genial mathematician called Blaschke (L. Santalo was one of his students) is to consider this kind of space as an "homogeneous space" https://en.wikipedia.org/wiki/Homogeneous_space . For example the space of quadrilaterals is, said in an approximate manner : $\mathbb{R}^8$ (4 points with 2 coordinates) quotiented by the transitive group of similitudes (in the sense of group actions, but once again, I advise you to have a look at Santalo's book, reviewed here https://projecteuclid.org/download/pdf_1/euclid.bams/1183539854).

2) I have been working in the past on similar issues, and I found a pertinent subdomain called centro-affine geometry also developed in the first part of XXth century.

3) Last but not least. See this very recent article : "Quadrilaterals in_Shape Theory II Alternative Derivations of_Shape Space. Successes_and_Limitations" by Edward Anderson https://arxiv.org/pdf/1810.05282.pdf that should bring you something (and even more...). The bibliography you will find therein has many self-citations... In particular you will see there is a part I and a forthcoming part III. I just discovered this article today : instead of taking days to analyse it at spare time, I preferred to share the information at once.

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  • $\begingroup$ Many thanks, Jean Marie, for the valuable hints! $\endgroup$ – Andreas Rüdinger Mar 17 '19 at 20:24
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    $\begingroup$ I just found this reference to your precise issue : "Coincidences of Centers of Plane Quadrilaterals" authors Al-Sharif, Hajja, Krasopoulos citeseerx.ist.psu.edu/viewdoc/… with Fermat point as the "fourth thief"... $\endgroup$ – Jean Marie Mar 17 '19 at 22:57

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