I once received a puzzle that can be described as follows: There are $12$ black pentagonal and $20$ white hexagonal pieces. The goal is to form a soccer ball from these (aka. truncated icosahedron). Similar to a (flat) jigsaw puzzle, each edge of each piece is equipped with either a "tongue" or a "groove" for interlocking (and there is only one type of these, that is, every tongue fits into every groove of every other piece). The assignment of tongues and grooves follows a highly symmetric pattern: Each piece has at least two tongues and at least two grooves; apart from that all possible shapes occur in equal number, that is:

  • For each of the four ways (up to rotation!) to pick two or three tongue positions for a pentagon, there are three puzzle pieces with that pattern
  • For each of the ten ways to pick two or three or four tongue positions for a hexagon, there are two puzzle pieces with that pattern.

It seems highly incidental that the required number of pieces is a multiple of the possible configurations - then again this is just the law of small numbers at work.

Anyway, I have spent hours on solving the puzzle manually, but never succeeded. The tricky thing is that it is very easy to just go ahead put together all but one or two pieces, but those last pieces just never fit. On the other hand, once I sat down and wrote a program to solve the puzzle by backtracking it began spatting out gazillions of solutions faster than the screen could scroll ... which is of course embarrassing for me as a puzzler.

My question is: Given the highly symmetric setup of the collection of puzzle pieces, there should be some "symmetric" solution to the puzzle. Can anyone find such a symmetric solution?

Of course, the term "symmetric" has to be taken a bit loosely here (and that makes this question somewhat opinion-based), after all the threefold repetition of pentagonal pieces and twofold repetition of hexagonal pieces excludes anything that is actually invariant under a symmetry movement of the soccer ball ($\gcd(2,3)=1$). So anything that can be found by contemplating instead of computer power and/or that can be described by a nice pattern instead of a piece-by-piece enumeration would be acceptable.

  • $\begingroup$ Just to make sure: is rotational (as opposed to reflectional) symmetry the only degree of freedom we can exploit. In other words, have the 5/6-gons got a clearly "preferred" side that should end up facing the viewer of the finished puzzle? $\endgroup$ – Jyrki Lahtonen Oct 13 '15 at 7:24
  • $\begingroup$ @JyrkiLahtonen Good point. With the pieces before my eyes I would not have thought one migh tconsider this. The pieces have an outer face because they look more or less like frusta (with the big, outer face spherical to make a fine round ball in the end) $\endgroup$ – Hagen von Eitzen Oct 16 '15 at 15:58

My solution (a point outside an edge means a tongue, inside, a groove):


It is not completely symmetric, but still has a lot of symmetry.

The idea of my construction is to fill a half of the ball, then "flip" the filling so that the other half gets filled in inverted manner.

Step 1. Start with the central hexagon of the Schlegel diagram. It takes alternating tongues and grooves.

Step 2. Now we assign the tongues and grooves to the belt between two halves of the ball. This has to be very symmetric, since we are going to flip our construction symmetrically. Each pentagon gets one tongue and one groove in the belt. The hexagons alternate: two tongues, two grooves, two tongues,...

Step 3. Now we fill the inner pentagons symmetrically. Well, almost. I didn't manage to achieve the complete ($60^\circ$-rotational) symmetry, so one of the pentagons near the belt got two grooves instead of three.

Step 4. Fill in hexagons having in mind that they will be inverted to the upper (e.g. only one hexagon with three consecutive tongues and three consecutive grooves is allowed). I did this manually; spent couple of hours trying to get this work. Finally, I decided to spoil the symmetry of pentagons a little, then after several tries found the solution.

Step 5. Invert the construction to the outer half (inverting means replacing tongues with grooves and vice versa).

Animation of the process:

enter image description here

  • $\begingroup$ This worked and seems to be reasonably symmetric (under the aforementioned restrictions that much symmetry could not be possible after all). The proof is in the pudding - or rather puzzling, which I successfully performed with my real life puzzle along this pattern $\endgroup$ – Hagen von Eitzen Oct 18 '15 at 9:33

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