One answer: Things don't inevitably match up on the bottom. The tetrahedron is the counterexample. Constructing it based around one vertex as you describe (ca

One answer: Things don't inevitably match up on the bottom. The tetrahedron is the counterexample. Constructing it based around one vertex as you describe (call it the north pole) does not result in a matching cap around the south pole. Of course, the gap around the south pole is still congruent to the three original triangles. It must be by the rules of Platonic solids, because the other three vertices must each behave identically to the original north pole, meaning the south-pole-centered triangle must be identical to the other triangles. Here's a more general answer. The solid always has N-way radial symmetry when viewed from above any vertex, such as our north pole. All longitude lines at 360°/N intervals are identical from this method of construction. There is nothing to differentiate the triangle of your icosahedron on the 0° meridian from its counterpart at 72°, so they will all fall into the same relationship with the south pole. The polygons can't overlap the south pole from one direction while falling short from another. In the tetrahedron, the south pole gets overlapped equally from all three radial directions. In the other four solids, the edges and faces meet exactly at the south pole. I can't answer exactly why that last sentence is true. It reduces to, why does this construction method always produce symmetry across the equator for the four non-tetrahedron solids? Your strip of 10 triangles on the icosahedron: if that is indeed equatorial, then we have both latitudinal and longitudinal symmetry which renders your north and south hemispheres identical so things will match up. But why does that strip of 10 triangles end up centered on the equator? I do not have the answer to that, but maybe someone does.