Additions to

Zome Geometry

George W. Hart   and  Henri Picciotto

Green Giants

Challenge: With 24 green struts, make a convex polyhedron consisting of eight identical equilateral triangles and six identical rhombi. Four struts meet at each vertex. The cuboctahedron would be a solution (because squares are rhombi) so to make it more challenging, use non-square rhombi.

The polyhedron illustrated above came as a real surprise to me.  It consists of twenty equliateral triangles and sixty identical rhombi, arranged with icosahedral symmetry.  At left is the Zome construction, which requires 150 green struts of any one size.  I liked it so much that I made a wooden model, about one foot in diameter, shown above at right. (If you want to make a wooden or paper model, the acute angle of the rhombi is about 66.14 degrees.)

I knew of other polyhedra with 20 equilateral triangles and 60 tetragons (e.g., the propello-icosahedron,) but they involved kites, trapezoids, or other less regular tetragons.  This rhombic version arose in joint research with Douglas Zongker.  It is actually the Minkowski sum of the icosahedral compound of five tetrahedra, and first appeared (as far as I know) as a computer model in our joint paper to appear in the Proceedings of the Bridges 2001 conference.  As the paper mentions: if several polyhedra are individually Zome constructible, then so is their Minkowski sum (in any of the 60 relative orientations in which their Zomeballs are aligned).  So I knew it was posible to build this with Zome, but working out the proper directions was still a bit tricky.

What is a Minkowski sum? you ask?  One answer is that if you find the (x,y,z) coordinates of each point inside of polyhedra P1 and P2, then the Minkowski sum of P1 and P2 is the set of all points you can get by summing any one point from P1 with any one point from P2.

If you try to make an analogous rhombic version of the propello-dodecahedron, the rhombi become coplanar in threes, forming the hexagons of the truncated icosahedron. For other propellorized polyhedra made with rhombi see Jim Mcneill's page of "rhombified stellations".
You can construct a more peculiar green polyhedron, with tetrahedral symmetry, if you take the Minkowski sum of just four of the five icosahedrally compounded tetrahedra.  It is easiest to build if you start with the above and remove four tetrahedrally arranged triangles plus the six "roads" of four rhombi that connect them.  This leaves four regions which fit together making a 96-edge polyhedron with 52 faces---16 triangles and 36 identical rhombi.

You can keep going down, and make the Minkowski sum of just three or two of these tetrahedra, but they have less symmetry.  The sum of three tetrahedra has 12 triangles and 18 rhombi.  The sum of two tetrahedra has 8 triangles and 6 rhombi.  Both have one 3-fold rotation axis and three 2-fold axes.  All come in left-hand and right-hand forms.

Challenge Answer: The last of these answers the challenge.  As scaffolding, make a flat starburst by inserting three b1's and three b2's into alternate holes in the yellow plane of a central Zomeball.  Balls at the ends of the blue struts can be connected with six g1's to make an irregular hexagon.  Rising above each b1 add two g1's to make a b1-g1-g1 isosceles triangle.  Connext their apexes with three g1's, forming an equilateral triangle parallel to the irregular hexagon.  Connect the triangle vertices to the hexagon vertices with six g1's that form three triangles and three rhombi.  Turn this over, remove the scaffolding (i.e., the central ball and struts it touches) and make an identical half on the other side of the hexagon.  The result is topologically equivalent to the cuboctahedron.