Does anyone know what is the fewest-piece dissection of the surface of a regular tetrahedron to the surface of a cube (of the same area)?

It is well-known that the volume of a regular tetrahedron cannot be dissected to the volume of a cube, because their Dehn invariants differ. But their surfaces have a dissection, by applying the Boyai-Gerwien theorem. Applying that theorem in one way (among several options) leads to an $31$-piece dissection (if I calculated correctly), illustrated below. But this is surely very far from optimal (and hardly aesthetically pleasing). Likely the question has been explored, but I have not found any literature. It could make a pleasing contrast to the impossibility of a volume dissection.

Let the cube edge length be $1$, so its surface area is $6$. A tetrahedron edge length of $L= 2^{\frac{1}{2}} 3^{\frac{1}{4}} \approx 1.86$ leads to a surface area of $\sqrt{3} L^2 = 6$.

          Surface dissection of regular tetrahedron to cube.

Related: "Covering a Cube with a Square."

  • $\begingroup$ You can avoid some cuts if you use folding. I would look at converting two folded triangles to a rectangle of dimensions 3 by 1. Even if you have to cut at the folds, this may involve fewer pieces. Gerhard "Thinking Bent Instead Of Fragmented" Paseman, 2017.02.24. $\endgroup$ – Gerhard Paseman Feb 24 '17 at 17:00
  • $\begingroup$ @GerhardPaseman: Yes, one can easily improve that dissection. Unfolding the tetrahedron and cube as single-piece nets, rather than treating each face separately, would improve the count. $\endgroup$ – Joseph O'Rourke Feb 24 '17 at 17:16
  • $\begingroup$ Also, in your original dissection, you can save on the piece count by only cutting two triangles in half, not cutting all four triangles. I think this change alone will reduce your piece count by 12. Gerhard "This Makes It Much Simpler" Paseman, 2017.02.26. $\endgroup$ – Gerhard Paseman Feb 26 '17 at 23:26

If folding is allowed, there is a 5 piece dissection: 3 pieces to reshape the rectangle from tetrahedral to cubical, plus two more pieces to arrange the latter into an arrangement that folds around a cube.

Start by cutting a tetrahedron along two disjoint edges plus an altitude of any face such that the altitude is perpendicular to one of the cut edges (and thus shares a vertex with the other cut edge). One unfolds to get a rectangle roughly 1.6 by 3.7. Now cut this into three pieces similar to the example in the question to get a 2 by 3 rectangle. If I am not mistaken, there is room in the corner of each large piece to cut out a unit square. The resulting 5 pieces can be folded around a unit cube.

Even if you have to cut the rectangle further to get the individual faces, I think one gets fewer than 20 pieces with this method.

Edit 2017.02.28 GRP : Here is my next attempt at explaining by picture. I hope it shows how a five piece folding dissection is achieved. Unfortunately, this version does not improve much on a 31 piece unfolded dissection.

End Edit 2017.02.28 GRP.

Gerhard "You're Going To Rewrap Anyway" Paseman, 2017.02.27. second attempt

  • $\begingroup$ Even if I am mistaken about the corner squares, it should be possible to take a 2x1 'bite' out of some corner to get a five piece total. Gerhard "Time Now For Some Coffee" Paseman, 2017.02.27. $\endgroup$ – Gerhard Paseman Feb 27 '17 at 17:18
  • $\begingroup$ Now that this part is available, similar constructions appear to me. Dissect the cube surface into two 1 by 3 rectangles. Arrange into a 2 by 3 or 1 by 6 rectangle as you please. Reshape into the tetrahedral rectangle, again with a five piece dissection. Score and fold around a tetrahedron. I am sure there is some software out there to help animate this. Gerhard "Putting Pictures In Other Minds" Paseman, 2017.02.27. $\endgroup$ – Gerhard Paseman Feb 27 '17 at 17:55
  • $\begingroup$ I am not sure about "there is room in the corner of each large piece to cut out a unit square," but I think the basic idea is correct. Very nice! I posted an image. $\endgroup$ – Joseph O'Rourke Feb 27 '17 at 21:33
  • $\begingroup$ If I count correctly, this will give a 23 piece dissection, with a few itty bitty pieces to manage. Gerhard "Please Don't Mix Them Up!" Paseman, 2017.02.28. $\endgroup$ – Gerhard Paseman Mar 1 '17 at 5:54

I think a good solution starts with rearranging two triangles into a 3 by 1 rectangle of the same area. In particular, start symmetrically: find the center $C$ of the rectangle, and align one of the triangles so it has an edge aligned with side of length 3 and $C$ is on its perimeter. Do a $180^\circ$ rotation to align the other triangle similarly. You now have the center square decomposed in 2 pieces, and a large portion of a triangle devoted to one of the remaining squares.

      (Added by JORourke.)
You now have the problem of rearranging two triangular pieces into the gap in the unfilled portion of the square which is a right triangular piece of similar shape. I suspect this corner portion can be done with 5 or fewer pieces, given a total for this half of 14, assuming the rectangle needs to be cut.

You can also offset this construction slightly, getting rid of one of the triangle snippets by covering it with the rectangle. This might reduce the piece count a bit.

Edit 2017.02.25 GRP Here is my first attempt at including an image representing part of the dissection. The main task is to assemble right triangle MNO from an equilateral triangle with one vertex at M and a snippet outside of MNO but sharing a vertex with O. I've started such a process, but I believe Joseph knows a more efficient way. Enjoy the picture.

End Edit 2017.02.25 GRP

Gerhard "Leaves The Illustrating To You" Paseman, 2017.02.24.

[1]: https://i.stack.imgur.com/IUy6R.jpgenter image description hereenter image description here

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    $\begingroup$ I added a figure for the first part of your construction, but I couldn't understand the remainder. Sorry! $\endgroup$ – Joseph O'Rourke Feb 24 '17 at 22:18
  • $\begingroup$ Then you have my permission to edit the text as well. Add a (rotated by 180) degrees copy of the triangle so that they adjoin. Let the parallels of the two bases be 1 unit apart. Make sure the triangles share a line segment. Center the configuration onto the rectangle. Gerhard "Not Quite The Verbal Painter" Paseman, 2017.02.24. $\endgroup$ – Gerhard Paseman Feb 24 '17 at 22:32
  • 1
    $\begingroup$ Clearly I have not mastered the "Add Picture" aspect of editing. Apologies to Joseph, and anyone who can repair it to include his graphic and the better of mine has my gratitude. Gerhard "Really Prefers Command Line Editing" Paseman, 2017.02.25. $\endgroup$ – Gerhard Paseman Feb 25 '17 at 19:42
  • $\begingroup$ It may be better to take the snippet from MNO, and then join the two quadrilaterals (remember the rotation!) to form a rectangle which then is dissected into two equilateral triangles of equal size. (Or four of them to make four.) In any case, the starting configuration should inspire some other dissections. Gerhard "Enough Mental Picturing For Now" Paseman, 2017.02.25. $\endgroup$ – Gerhard Paseman Feb 25 '17 at 21:19

Here is a dissection based on Gerhard's latest idea. I didn't add this directly to his post because it is not identical to his $5$-piece dissection from unfoldings.

The tetrahedron is unfolded to a $3.722 \times 1.611 = 6$ rectangle. The cube is unfolded to two $3 \times 1$ rectangles, which are stacked to $3 \times 2$. This leaves a $0.388 \times 3 = 1.164$ rectangle to cover a $0.722 \times 1.611 = 1.164$ rectangle. This is accomplished by the magic of the Bolyai-Gerwien proof.

This appears to be $5$ pieces to form the $3.722 \times 1.611$ rectangle from the cube's two $3 \times 1$ rectangles. I consider this a solution, starting from surface unfoldings, and have accepted Gerhard's answer.


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