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When doing certain things with polyhedra, like applying Conway's operations or developing frame models for 3D printing, it's helpful to know whether all the faces face directly away from the center of the polyhedron. In other words, would a line passing through the center of the polyhedron and the center of each face be perpendicular to that face?

I've looked through terms for characteristics and symmetries of polyhedra such as here and here, and I don't see a term for that characteristic. For brevity, let's call this characteristic Y for purposes of this question.

  • regular: too narrow. Archimedean solids appear to be Y but not regular.
  • orthohedral: This set of articles defines orthohedral as having "all faces... equidistant from the center." I'm pretty sure this is too narrow as well, because it would exclude an Archimedean truncated tetrahedron.
  • uniform: defined as having "regular polygons as faces and is vertex-transitive". This is closer, but again, I'm pretty sure it's too narrow. For example, the pseudorhombicuboctahedron (a Johnson solid) is Y but isn't uniform. A simpler example of a non-uniform Y polyhedron would be a rectangular cuboid.
    • Archimedian: If defined as a subset of uniform polyhedra, this is also too narrow. Even if defined more weakly to include pseudo-uniform polyhedra, it excludes most rectangular cuboids.
  • Catalan solid: a brief look at these suggests that they may all be Y. (But Catalan solids don't include all Y polyhedra.)
  • Johnson solid: Too broad (and too narrow, unless it's defined as including uniform polyhedra). Plenty of Johnson solids, such as J7, are not Y.
  • canonical: All edges equidistant from the center. This isn't true for rectangular cuboids.
  • circumscribable / equiradial: (Thanks to @OscarLanzi for suggesting this direction.) A polyhedron whose vertices all lie on the surface of a sphere, i.e. equidistant from the sphere's center. But the center of the sphere might not be what you expect as the center of the polyhedron... it might even lie outside the polyhedron.
    • Which raises the point that in order to answer this question, one has to first specify what "center" of the polyhedron one has in mind as well as what we mean by the center of each face. Using circumcenter for both might give us a well-defined way to answer this question: Y is equivalent to circumscribable (if that's always true).
    • So far, I have been using the arithmetic mean of the vertices as the center of each face and each polyhedron. That seems to work out fine for (true or pseudo) uniform polyhedra. And it has this advantage: For an operation like expansion, in which the faces "are separated and moved radially apart," the center that faces are moving away from would have to be inside the polyhedron. Which isn't always true of the circumcenter.

So, the question is, is there a term for Y (for some reasonable definition of 'center' of face and polyhedron)? If not, is there a term for an equivalent characteristic, i.e. a characteristic that is true for a polyhedron iff Y is true?

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    $\begingroup$ Observation: all polyhedra inscribed in a sphere have a normal from every facial plane passing through the center of the sphere. The appropriate normal would be from the circumcenter of its respective face (the faces would all be inscribed in circles), which could be outside the area of the face proper. $\endgroup$ Oct 5, 2021 at 15:28
  • $\begingroup$ @OscarLanzi: Thanks for this observation, especially regarding the circumcenter of the face. I hadn't thought of the appropriate 'center' of a face being something other than the mean of the vertices. I believe that polyhedra inscribed in a sphere are what numericana.com/answer/polyhedra.htm#Equimetric call equiradial polyhedra. So maybe equiradial polyhedra are Y, but are all Y polyhedra equiradial? $\endgroup$
    – LarsH
    Oct 5, 2021 at 15:34
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    $\begingroup$ @OscarLanzi Your point about the circumcenter of the face, which could be outside the face proper, leads me to reconsider what I mean by the center of the polyhedron. I said in my question that J1 is not Y, and yet it could be inscribed in a sphere. So maybe J1 is Y, if the center of the polyhedron is the center of its circumscribed sphere! $\endgroup$
    – LarsH
    Oct 5, 2021 at 16:01

2 Answers 2

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As discussed in comments: if a polyhedron is inscribed in a sphere (equiradial), then a normal line to each facial plane passes through the center of the sphere. The appropriate normal to each face would be through the circumcenter of that face (all faces of an equiradial polyhedron are inscribed in circles). Note that the circumcenter of the polygon may lie outside the polygon proper, and the center of the sphere in which the polyhedron is inscribed may likewise lie outside the volume enclosed by the polyhedron.

In such a case, if we define centers as circumcenters or their higher-dimensional equivalents, then all equiradial polyhedra are Y. However, for other polyhedra this type of center is not defined for either the faces (generally) or the polyhedron, so this result cannot be extended to classify other polyhedra.

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    $\begingroup$ Thanks for making this helpful result into an answer. I will accept it if no better answer comes along soon. You stated that all equiradial polyhedra are Y. I assume you intentionally are not claiming that all Y polyhedra are equiradial, i.e. that equiradial is equivalent to Y? $\endgroup$
    – LarsH
    Oct 6, 2021 at 10:54
  • $\begingroup$ Upon further reflection ... If we define center in the way you described, then no polyhedron can be classified as Y unless it has a circumcenter, which requires that it be equiradial. So the terms equiradial and Y would in fact be equivalent. $\endgroup$
    – LarsH
    Oct 6, 2021 at 13:18
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If you forget about the "center" of each face (which as you've noted, is hard to define) and only ask for lines passing through each face, orthogonal to the face, and all meeting in one point inside the polytope, that's called a Rogers center.

Definition. A Rogers center of a polytope is an interior point $\mathbf{o}$, fixed by any symmetries of the polytope, such that the orthogonal projection of $\mathbf{o}$ onto the hyperplane affinely spanned by each facet is within the relative interior of that facet.

This is the most relevant concept I know that doesn't require the vertices to lie on a sphere. So you'd say the polyhedra you want "have a Rogers center" or maybe "are Rogers-centered" (I don't know if anybody says that.)

Two terms that might be handy for the circumscribable case:

Definition. A polytope $P$ is well-centered if all its vertices lie on a sphere, and the center of this sphere is in the relative interior of $P$. $P$ is completely well-centered if every face (including $P$ itself) is well-centered.

I took these terms from a PhD thesis I found (Hirani, A.N., 2003, "Discrete exterior calculus") and used them in my own thesis (2015, "Convex Polytopes and Tilings with Few Flag Orbits"). I wrote a couple of blog posts discussing some of the same problems you bring up: Well-centered-ness and Well-centeredness and Orthoschemes.

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  • $\begingroup$ Thanks for the pointers to these concepts and blog posts. I'll take a look. $\endgroup$
    – LarsH
    Feb 6 at 11:17

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