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)