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I noticed usual polyhedra have some vertices joining exactly 3 edges or some triangular faces (or both).

Out of curiosity I started wondering if there is a polyhedron with the following constrains:

Each vertex must join at least 4 edges and

each face must have at least 4 edges

I have looked through any number of random examples on the internet and none of them seem to fulfill these conditions, but yet I don't know if such a body is possible at all.

Rol
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2 Answers2

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@joriki's answer demonstrates that there is no polyhedron of genus 0 satisfying the stated conditions, which leads one to wonder whether it's possible to satisfy the conditions with a higher-genus polyhedron. And in fact, it is. One can construct a toroidal polyhedron with 9 quadrilateral faces (four meeting at each vertex), 9 vertices, and 18 edges. Basically, you obtain this if you take three equilateral triangular prisms, miter both ends of each at 30 degrees (so that when they are joined, they will make a 60 degree angle) and glue them all together in an overall triangular fashion. Here's a picture:

toroidal polyhedron all faces quadrilaterals, all vertices degree 4

(This same picture also appears in the answer to question #1005105.)

Community
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Glen Whitney
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I'll answer the first question. (Asking more than one question per post is discouraged.)

Since each edge is incident at two vertices and adjacent to two faces, your conditions imply

$$e\ge\frac{4v}2=2v\;,$$ $$e\ge\frac{4f}2=2f\;,$$

where $v$, $e$, $f$ are the numbers of vertices, edges and faces, respectively. Adding the two inequalities yields $e\ge v+f$, which contradicts Euler's polyhedron formula, $v-e+f=2$.

joriki
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    Thanks, I split the question into two. – Rol Sep 08 '15 at 13:10
  • The calculation suggests that there might be a polyhedron as described of higher genus, and indeed that is the case: It is easy to construct a toroidal polyhedron with 9 quadrilateral faces, four meeting at each vertex, 9 vertices, and 18 edges. (Basically take three equilateral triangular prisms, miter them at 30 degree angles at both ends, and glue them together.) – Glen Whitney Sep 20 '18 at 22:39
  • @GlenWhitney: Thanks, that's interesting. But isn't it $60°$ at both ends, to make up $360°$ in total? – joriki Sep 21 '18 at 01:32
  • A 30 degree miter one each of two pieces of stock creates a 60 degree angle between them when you glue them together. (Think about when you want to make a picture frame, you cut each piece of the framing at 45 degrees.) And three 60 degree angles makes the toroid close up with three sections into the overall shape of something like a picture frame in which you could put an equilateral triangular picture. Actually, there's an image (https://i.stack.imgur.com/taTrs.png) of this polyhedron in question #1005105, sorry I didn't look before. – Glen Whitney Sep 21 '18 at 11:26
  • @GlenWhitney: I see, thanks -- I misunderstood what you mean by a $30°$ miter. Actually, with that nice picture, I think your comment would make a nice answer to this question in its own right :-) – joriki Sep 21 '18 at 11:41