- Puzzle Zapper Blog - http://puzzlezapper.com/blog -

Flexible pentominoes on rhombic polyhedra

If you subdivide the faces of a rhombic triacontahedron [1] into 2×2 grids, you can tile the polyhedron with two copies of each pentomino.

One way of looking at this figure is as a tiling of the projective hemi-rhombic triacontahedron. The projective (also known as abstract) polyhedra can be formed by identifying the opposite faces of certain polyhedra with each other. So the projective hemi-cube has three square faces, and the projective hemi-rhombic triacontahedron has 15 rhombic faces. Stitching together the opposite sides of the unshaded area in the figure is a way to form this 15 face “polyhedron”.

I came up with that one a couple of years ago, but I neglected to put up a blog post because I didn’t like the graphic enough. I suspect that it’d look really cool if the lines of the rhombic triacontahedron were properly projected onto a flat disk, but I don’t have the expertise to make that happen. I finally decided that it was worth sharing even if it doesn’t look as cool as it could.

Below is another tiling of subdivided rhombi. The significance of this figure is that four copies could be used to cover a rhombic hexecontahedron [2].