Existence of Platonic and Archimedean solids Which combinations of regular polygons around a corner are candidates for Platonic and Archimedean solids can be decided locally. (For a corner of a polyhedron we need at least 3 faces and for a convex polyhedron the sum of the angles of polygon corners must be less than 360°. These two restrictions imply 5 possible combinations of equal regular polygons - the basis for the 5 Platonic solids.) The choice of the regular polygons around a corner defines the angles between the faces.
Is there an argument (simple or elaborate) why all these candidate combinations can actually be continued to a convex, non-self-intersecting polyhedron - without having to construct each individual solid?
 A: Since the polygons are regular and congruent, can you argue that if three or more can form the vertex of a solid angle with one of their plane angles, then by symmetry they can do so with all of them, and thereby form a convex regular solid? Any one corner implies neighboring corners. This simple argument rests on something analogous to the principle of isotropy. Might it work?
A: Here are a couple limitations you can use to narrow down your "search":
1)  Sum of angles at each vertex:
Suppose you add up all the angles of all the polygonal faces.  If there are $F$ faces and $E$ edges then the combined sum is $180°×(2×E)-360°×F$ (the first term comes from $180°$ times the total number of sides of all the polygons, the sum counting each polyhedral edge twice; the second term is the $360°$ you normally subtract for each polygon).  This simplifies to $360°×(E-F)$ which, because of the Euler characteristic, equals $360°×(V-2)$ where $V$ is the number of vertices.  Then the sum of angles at each vertex must be $360°$ minus a divisor of $720°$.
For instance, you might suppose that a solid could be formed with a square and two triangles at each face, but ... no.  The angles at each vertex add up to $90°+60°+60°=210°$, but $360°-210°=150°$ does not divide $720°$.
2)  Fitting around odd-sided polygons:
When you have a face with an odd number of sides, all faces adjoining it along the edges must be congruent.  Suppose you try to make a polyhedron with two pentagons and a hexagon at each face.  The vertex angles add up to $360°-24°$ and $24$ divides $720$, so the coast is clear. You adjoin a second pentagon to one edge of the pentagon.  Then you apply a hexagon to the next edge, followed by another pentagon, a second hexagon, then the fifth edge will be joined to ... oops.  You see that neither of your choices works because of the odd number of sides on the pentagon you're trying to surround.  Instead you'd have to alternate your pentagons with more pentagons (forming a regular dodecahedron) or hexagons with more hexagons around a single pentagon (forming an old-style soccer ball polyhedron).  
You still get infinitely many possibilities.  To get Archimedes' finite list you have to arbitrarily (and to me, dubiously) reject prisms and polyhedra with too many triangles.
Have fun!
