A cube has 6 faces and 12 edges, and 6 conveniently divides 12, so can a cube be covered by 6 congruent copies of an object composed of a square face and 2 edges? Equivalently, draw the letter L on each face so each edge has exactly one leg of an L parallel and close to it. Or solder 6 Ls to form a wireframe cube. Yes, this can be done. Along the equatorial belt, 2 right side up followed by 2 upside down. The polar caps are then forced.
The covering can also be done where the two edges are the opposing sides of the square face. Equivalently, draw an equals sign = on each face with the same constraints as above with L. Opposing faces end up having the same orientation, which is nicely symmetric.
Regular tetrahedron: 4 faces and 6 edges, so 1.5 edges per face, so not possible. (Maybe it's possible with partial edges.) Similarly, octahedron 8F 12E, dodecahedron 12F 30E, icosahedron 20F 30E, none divide evenly.
Square tiling: 2 edges per face. Hexagonal tiling: 3 edges per face. These can be done by (for example) drop shadow.
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