What series of 'hyperpolyhedrons' do exist? Is there an effective way to derive their cross-sections by 3-d subspace?

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There are two obvious series of 'hyperpolyhedrons'.

  • 'Hyperoctahedron' with vertices $(\pm1,0...0), (0,\pm1,0,...0)...(0,...0,\pm1)$ and each vertex connected by an edge with each other vertex except its opposite

  • 'Hypercube', dual to hyperoctahedron, with vertices $(\pm1,\pm1,\pm1,...\pm1)$, with edges connecting vertices that differs by sign of exactly one coordinate.

Probably, a series of "hypertetrahedrons" should exist, with 'hypertetrahedrons' containing number of vertices equal to number of dimensions of 'hydpertetrahedron's' native space plus one, though not aware of a simple way to get coordinates of their vertices.

Questions:

  • Are their series of 'hypertetrahedrons', 'hypeicosahedrons' and 'hypedodecahedrons' ? How to obtain coordinates of their vertices?

  • Is their a general and reasonably efficient algorithm to get a list of all vertices, edges and faces of a section (not projection) of any 'hyperpolyhedron' mentioned above with a 3d subspace of its 'native' space?

  • What are the proper words for all above?

  • Are there any entry-level easily available (in form of pdf or html page) books/articles on the web? Could you give a link?

upd. Yes, I'm asking about regular, convex 'polyhedrons', or, as said in an answer, polytopes.

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They are called regular polytopes. As you said there are three series of regular polytopes in all dimensions: hypercubes, their duals (which are the generalized octahedrons), and simplexes (generalized tetrahedra). For dimensions greater than $4$ these are the only ones if we also assume they are convex. In dimension two we have the regular polygons, one for each $n$. In dimension three we have the Platonic Solids. In dimension four there are six regular convex polytopes.