Currently on my workbench – an exploded icosahedron

So that left me wondering what the dihedral angle is between a triangle and a square in this shape. Then I saw that if I exploded the icosahedron in a similar way, I would again end up with a rhombicosidodechahedron! So that dihedral angle must be ½ the dihedral of the icosahedron (138º11'/2 = 69º6') + the 90º of the square = 159º06'.

I presume there is some more elegant way of calculating these angles; however, my math abilities just aren't at that level, so I'm extremely happy to have discovered this roundabout method. I feel like I own the rhombicosidodechahedron now, and I look forward to fashioning a lamp in that shape someday.

In fact, I want to make a few lamps based on that shape; I also really like that shape when it is open at the top, with just the base decagon and the five lower overlapping decagons on the sides. That shape looks like a flower bud opening.

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