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Start with five identical cubes:

five identical cubes

Your challenge:

  1. Cut and unwrap all five cubes into five identical cube nets.
  2. Show how to re-fold these five cube nets to form the surface of a single larger cube, with five times the surface area of one of the original cubes.

Rules and clarifications:

  • The original five cubes must be the same size.
  • The cube nets must be made by cutting the original cubes along their edges only.
  • The five cube net shapes must be identical to each other. (Mirror images do not count as identical.)
  • The five cube nets must cover the entire surface of the final cube, with no gaps and no overlaps.
  • The edges of the cube net shapes do not need to line up with the edges of the final cube.
  • I am aware of one solution, but I can't guarantee it's the only one. Any solution that meets the above requirements is valid.

Note:

This is a variation on my previous puzzle: Turn two cubes into one! This new puzzle is very similar to that one in terms of how the puzzle is worded, but the solution is quite different, and was (for me, at least) much harder to find. (Honestly, when I started exploring this one I was kind of expecting to make a puzzle where the challenge is: "Find a clever way to prove this thing is impossible".)

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  • $\begingroup$ Suppose the small cube nets are painted on one side. Must the large cube show only painted faces, and must the nets be identical excluding mirror images when the painted side is up for all nets? $\endgroup$ Nov 14, 2022 at 11:48
  • $\begingroup$ "The edges of the cube net shapes do not need to line up with the edges of the final cube." This rule is unnecessary as you can not form a square face with 5 smaller faces. $\endgroup$
    – Minot
    Nov 14, 2022 at 16:00
  • $\begingroup$ @ParclyTaxel Yes to both. If you choose a cube net for which 'handedness' is meaningful (where a mirror reflection would look different), then all of them must have the same handedness as they appear on the larger cube. $\endgroup$ Nov 14, 2022 at 17:43
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    $\begingroup$ @Minot I agree the rule is not strictly necessary in order to define the puzzle. It's less of a requirement and more of a clarification. I just wanted to make it crystal-clear that rule 2 (cut on edges only) only applies to the smaller cubes, because otherwise (as you observe) the puzzle would not make sense and would be impossible. $\endgroup$ Nov 14, 2022 at 17:52
  • $\begingroup$ Would you like something harder? See my question. $\endgroup$ Nov 17, 2022 at 8:05

1 Answer 1

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Only one of the 11 normal cube nets allows the large cube to be formed – the cross-shaped one – and the solution is unique: The centre cell of each cross is marked with thin lines above.

The solution was found and proved to be unique using an exact cover program on the tiling of the larger cube, which has to have side length $\sqrt5$. I started with six "rings", where each ring contains the squares visited if you start from some square and keep moving forwards: From this I found for each one-sided net (as required by the problem rules; the large cube tiling is chiral) all admissible ways of placing it on the large cube tiling, using a turtle graphics-like approach to specify the cube nets (aided by the "ring decomposition" of the squares from above). These ways in turn form the input for the corresponding exact cover problem.

All nets but the cross-shaped one led to zero solutions; the cross-shaped one led to 24. Since the tiled large cube has 24 symmetries and the solution above is totally asymmetric, however, the above solution is unique.

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  • $\begingroup$ Did you use a specific program to come up with/draw such a thing? It looks beautiful :) $\endgroup$
    – Minot
    Nov 15, 2022 at 13:18
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    $\begingroup$ @Minot Inkscape to draw out the solution. See my expanded answer. $\endgroup$ Nov 15, 2022 at 13:23
  • $\begingroup$ Very impressive! I strongly suspected my solution was probably unique, but was unable to prove it with the methods I was using (painting faces by hand in a 3D modeling program using trial and error). And thank you for sharing your code with us. $\endgroup$ Nov 15, 2022 at 17:09

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