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So I've been playing around with polyhedra for my own amusement, but I ended up with some that I couldn't find names for. I have been trying to find them on my own by Googling for polyhedra with these specific parameters and using Wikipedia and other sites, but the results have been more frustrating than not (I can find ones close to them but not them exactly). While I'm aware Greek is used in nomenclature for polyhedra, I don't really know all the other factors and don't know what these would look like.

Here are the parameters for them:

Vertices---Edges--Faces

22------------36-------16 tetrahedron + dodecahedron

26------------42-------18 cube + dodecahedron

28------------44-------18 cube + dodecahedron

24------------42-------20 dodecahedron + octahedron

26------------44-------20 dodecahedron + octahedron

18------------44-------28 icosahedron + octahedron

16------------42-------28 icosahedron + octahedron

I apologize if this is somehow unanswerable. I am not a mathematician; I really only understand the very basic equation for creating polyhedra (V - E + F =2). I know nothing about convex, concave, or, really, any other factors. I just assumed that there was something simple I was missing because of my lack of experience in this particular area.

For all of them, I had thought of adding together the faces of Platonic solids and trying to get a new polyhedra as a result. For instance a tetrahedron (4V 6E 4F) + a dodecahedron (20V 30E 12F) I ended up with something that could have 16 faces and either 22V and 36E or 24V and 38E. I assumed that because a hexadecahedron (which has 24V, 38E, and 16F) exists there must also exist a named polyhedron that has 22V, 36E, and 16F.

The same applies to the others (I listed which ones I was trying to combine next to each).

I'll try to be more specific, but please remember I really only know the formula and nothing else, so it seemed to me I could do most anything with it. Here is what was going through my mind:

So a tetrahedron has 4 vertices, 6 edges, and 4 face and a dodecahedron has 20 vertices, 30 edges, and 12 face. If I added their faces together 4 + 12 = 16. So I'd have some polyhedron with 16 faces. If I add their vertices 4 + 20 = 24. So I'd have some polyhedron with 24 vertices. If I add their edges 6 + 30 = 36. So I'd have some polyhedron with 36 edges.

Well, the formula is V - E + F = 2. 24 - 36 + 16 = 4: so I can't have a polyhedron with 24 vertices, 36 edges, and 16 faces. Well, what can I have?

Well, if I have 24 vertices, 38 edges, and 16 faces 24 - 38 + 16 = 2. That works! Is there a polyhedra that has 24 vertices, 38 edges, and 16 faces? -does a Google search- Yes there is! It's called a hexadecahedron!

If I can have only edges or vertices added, and I already found for 24 vertices and 16 faces, then what would I need for the edges? Well 22 - 36 + 16 = 2. I wonder what polyhedra has 22 vertices, 36 edges, and 16 faces. -tries to do a Google search. Cannot find. Searches Wikipedia through all Archimedean, Catalan and Johnson solids. Cannot find.-

Weird I can't find it. Maybe someone who knows math a lot better than me will be able to tell me the answer.

I'm sorry if I'm being a pain. I thought this would be simple. I can't emphasize enough that I really don't get how this works, exactly. I don't even know how to properly phrase all this. I'll try to clarify further with an example of another couple Platonic solids I added:

Tetrahedron + Icosahedron

4V, 6E, 4F + 12V, 30E, 20F = 16V, 36E, 24F (16 - 36 + 24 = 4)

14V + 36E + 24F = 2 (kisoctahedron; tetrahexahedron) 16V + 38E + 24F = 2 (disphenocingulum)

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    $\begingroup$ Are your polyhedra convex? Semi-regular in some sense? One can create irregular polyhedra with specific $V,E,F$ combinatorics that would never have been assigned a name. $\endgroup$ Mar 26, 2017 at 1:25
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    $\begingroup$ Seconding that - the information you've given just isn't enough, even granting convexity, to get unique information. Can you give nets for any of these polyhedra? Can you say where they were derived from? $\endgroup$ Mar 26, 2017 at 1:27
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    $\begingroup$ Comment to address the update: what you have aren't polyhedra without a specification as to what you mean by 'adding together the faces'; it's just a set of numbers. For instance, the dodecahedron isn't the only polyhedron with 20 vertices and 12 faces (and thus of necessity 30 edges); you can get numerous others by modifying the 'net' of the dodecahedron suitably. $\endgroup$ Mar 26, 2017 at 3:25
  • $\begingroup$ If you glue together two polyhedra along a pair of congruental faces, you must not forget to reduce the total number of faces by two and, if those faces have k edges each, the number of edges of the resulting polyhedron also must be reduced by k. $\endgroup$ Mar 26, 2017 at 11:20

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If you permit faces to pass through one another, as with the Kepler–Poinsot polyhedron, and don't count the "false" edges and vertices, then the counts $(V,E,F)=(22,36,16)$ —the first on your list— can be achieved by a polyhedron that has $f_6=8$ hexagon and $f_3=8$ triangle faces: $3 f_3 + 6 f_6 = 72 = 2 E$. It has been called the truncated stellated octahedron:


Tisso
Image from www.polytope.net.


In general, the problem of determining which $(V,E,F)$ triples are realizable as convex polyhedra is difficult, listed among Croft, Guy & Falconer's Unsolved Problems in Geometry:


CroftGuy
p.68 in: Croft, Hallard T., Richard K. Guy, and Kenneth J. Falconer. Unsolved Problems in Geometry. 1991.


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