Why can't a semi-regular solid have three different faces meeting at a vertex? I'm trying to establish why a semi-regular solid can't have three different faces meet at a vertex, for example a triangle, square and octagon.
I know its something to do the the internal degrees of each shape, $60°$, $90°$ and $135°$, but can't figure out how this is relevant, as $60°+90°+135°=285°$ and $285° \lt 360°$.
 A: There are two criteria you must satisfy beyond the vertex angles summing to less than $360°$. These criteria end up limiting you to two cases that actually exist.

*

*You have to have precisely the right vertex-angle sum. Just any old sum less than $360°$ per vertex doesn't cut it. Imagine pumping up the faces so they fit on the sphere in which the polyhedron is inscribed. This increases the vertex angles by an amount called the "spherical excess". With a cube, for instance, you start with three $90°$ angles at each vertex and after pumping it up, you have three $120°$ angles and thus you have generated $90°$ of spherical excess per vertex. With eight vertices this adds up to $720°$ spherical excess around the whole sphere. Well, for a whole sphere this spherical excess is always $720°$, so the spherical excess per vertex must divide that quantity. Your case of $360°-285°=75°$ doesn't do it.


*You have to pick the proper count of sides. Suppose you have three distinct regular polygons A, B, C meeting at each vertex. From a vertex follow the shared edge between, let us say, the A ploygon and the B polygon. Then to have semiregularity you must have a C polygon coming in at each end. Going all the way around the A polygon, the polygons meeting it must follow the alternating pattern B C B C... . If A, B, C are really all different, then this leads to a contradiction unless A has an even number of sides, and similarly B and C must also have even numbers of sides.
When we apply both criteria we find only a few surviving candidates. Three hexagonal vertices take up exactly $360°$ so to get a sum that's less, you need one of your even-sided polygons to be a square. Then if the other two polygons were both octagons we would again have the vertex angles adding up to a full $360°$, so a second polygon at each vertex has to be a regular hexagon. Now with these limited possibilities we can test various side numbers for the third polygon:

*

*Square-hexagon-octagon: Vertex angles add up to $345°$, thus $15°$ spherical excess per vertex. This is consistent with (1) if the polyhedron is given $48$ vertices, so there will be six octagonal, eight hexagonal and twelve square faces. If the octagonal faces are extended until their planes meet each other rather than the intervening faces, they will form a cube. Similarly extending the hexagonal faces yields a regular octahedron.


*Square-hexagon-decagon: Vertex angles add up to $354°$, thus $6°$ of spherical excess per vertex and we need $120$ vertices to cover the sphere. This implies twelve decagonal, twenty hexagonal and thirty square faces. Similarly to the previous case, extending the decagonal faces will produce a regular dodecahedron, while extending the hexagonal faces yields a regular icosahedron.


*Square-hexagon-dodecagon: Vertex angles add up to $360°$, so we are out of possibilities.
As you may surmise from my rather detailed description, both of these Archimedean solids actually exist. Go ahead and make them!
