Alan Schoen’s Tetrons (Cubons IV and Solitaire XVII)

After Alan’s cubons, now his tetrons. There is no end to it…

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They are constructed like the cubons: Take a tetrahedron, divide each edge into fifths, pick one of the four inner subdivision points on each edge, connect them to face centers and tetrahedron center to decompose the tetrahedron into four tetrons. Up to motions, each tetron is determined by the choice of three numbers from 1 to 4 (up to cyclic permutation), and, as for the cubons, there are 24 possible choices. In fact, the tetrons are just squished cubons.


The natural question for anybody obsessed with puzzles is whether these 24 tetrons can be used to assemble six tetrahedra. The answer is yes, you can group them in 11417 different ways into six sets of four so that this is possible. Above is one of them, and it is clear that this image is lacking, so below is the same solution, unfolded into nets. You will recognize what I discussed earlier as trions (albeit there with fewer subdivision points):

Sampletetronnet The tetrons are computationally much simpler than the cubons. For instance, we can again separate the 24 tetrons into 8 chiral and 16 achiral ones. Surprisingly, the 16 achiral ones can be assembled into four tetrahedra in exactly five different ways (up to rotations). Here they are, unfolded:


For the 8 chiral ones, the situation is a bit more complicated. There are two ways they can be grouped into two sets of 4, and in each case, there are two ways to assemble each set into tetrahedra. If we denote a single tetron by a list if three numbers that give the number of the chosen subdivision point as seen from the tetrahedron vertex of the tetron, then the partition of the 8 tetrons for the first solution can be denoted like so: {(1, 2, 3), (4, 1, 3), (2, 1, 4), (3, 2, 4)} and {(1, 2, 4), (4, 2, 3), (1, 4, 3), (3, 2, 1)}. Prettier are the nets. The first way to assemble them into tetrahedra is on the left, the second on the right.


And here is the other partition: {(1, 2, 3), (4, 2, 3), (2, 1, 3), (3, 2, 4)} and {(1, 2, 4), (4, 2, 1), (1, 4, 3), (3, 4, 1)}.

Again there are two different ways to assemble each quartet of tetrons into tetrahedra. I’ll show the solution next week.

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