Why semiregular polyhedra exists

Question

Yes. There are already many explanations for those five platonic solids and many other semiregular polyhedra such as Archimedean and prisms and antiprisms. My question is why they all exists. For example, if we want to find a regular polyhedra with faces consist of equilateral triangles, we may first specify the number of triangles at each vertex, say 3, 4, 5 and then by means of many methods e.g. Euler characteristics and double counting, can find the total number of faces and so on. And analyzing the angles between faces, one can prove that those cases are all feasible. Now there arose a question. Why that all the cases are feasible? E.g. it may be the case that tetrahedron and octahedron exist but icocahedron, during actual construction we may met a kink at the last stage so that it may not be able to be formed. However, this kind of situation does not happen. For every combination of expected combinations of regular polygons, we have almost always a corresponding really existing semiregular polyhedra. Can we give a explanation for existance of those combinations without invastigating individual cases? At my opinion, this is not trivial. For example, suppose that we want a polygon with faces given non-regular triangle with prescribed rules at each vertex how they are linked. Can we expect resulting solid almost surely exists? At my opinion, for this case, the answer would be 'no'.

Answer 1

"One can prove that those cases are all feasible."

Actually, no. For example, a 5-5-3-3 vertex that is seen in the pentagonal orthobirotunda does not occur in a semi-regular polyhedron. Peruse the Johnson solids and make a chart of all the regular polygon vertex types which cannot be realized.

Here is a disproof that the 5-5-3-3 polygon arrangement could occur in a semiregular polyhedron. Label one of the pentagonal faces $ABCDE$. Say that two pentagonal faces are shared along an edge $AB$. Then the remaining polygons around $B$ are both triangles including the face sharing edge $BC$. Thus the pentagonal face shared with edge $AB$ is alternated with a triangular face shared with edge $BC$. For the polyhedron to be semiregular this alternation must continue around the edges of $ABCDE$. Thus $CD$ is shared with a pentagonal face, $DE$ with a triangular face, $EA$ with a pentagonal face, and ... $AB$ is shared with a triangular face. Uh-oh, $AB$ was set up as shared with a pentagonal face. The assumed existence of a semiregular polyhedron is contradicted.

With a different ordering of faces -- 5-3-5-3 -- each pentagonal face shares edges only with triangles and vice versa. The contradictory alternation cited above is avoided, and a semiregular polyhedron can be and is formed with the 5-3-5-3 face arrangement.