# Covering a Cube with a Square

Suppose you are given a single unit square, and you would like to completely cover the surface of a cube by cutting up the square and pasting it onto the cube's surface.

Q1. What is the largest cube that can be covered by a $1 \times 1$ square when cut into at most $k$ pieces?

The case $k=1$ has been studied, probably earlier than this reference: "Problem 10716: A cubical gift," American Mathematical Monthly, 108(1):81-82, January 2001, solution by Catalano-Johnson, Loeb, Beebee.

(This was discussed in an MSE Question.) The depicted solution results in a cube edge length of $1/(2\sqrt{2}) \approx 0.35$.

As $k \to \infty$, there should be no wasted overlaps in the covering of the 6 faces, and so the largest cube covered will have edge length $1/\sqrt{6} \approx 0.41$. What partition of the square leads to this optimal cover?

Q2. For which value of $k$ is this optimal reached?

I have not found literature on this problem for $k>1$, but it seems likely it has been explored. Thanks for any pointers!

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I wonder if anyone in the packaging industry has the answer. –  Steven Gubkin May 4 '12 at 0:37
I believe there was a "Mathematical Games" column on dissections which had a Greek cross rearranged into a square. Perhaps that article also mentioned this problem? Gerhard "Testing Your Martin Gardner Fu" Paseman, 2012.05.04 –  Gerhard Paseman May 4 '12 at 15:51

You can cut a $\sqrt{6}\times\sqrt{6}$ square into 24 pieces that then cover the $1\times1\times1$ cube. Two triangles from the figure below plus one parallelogram make up one $1\times1$square. parts of pieces sticking out to the left can obviously fit back in the right, so 18 pieces, plus 6 parts sticking out equals 24. You can improve on this by stitching pieces across the cube edge to make one bent piece and by stitching some of the parallelograms back to the triangles.

[Added by O'Rourke:] Just to make Yoav's construction more explicit, here is how two triangles and a parallelogram fit together to form a $1 \times 1$ square:

[Added by Kallus:] Here's an illustration of a construction similar to Fedja's construction but with only five pieces. The first figure is the $\sqrt{6}\times\sqrt{6}$ square. The second is the $2\times3$ rectangle, which we fold into a cube by taking away the two yellow squares, folding the remainder, and adding the squares as the two missing faces.

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Brilliant!! :-) –  Joseph O'Rourke May 3 '12 at 23:51
Actually, any two polygons of the same area are equidecomposable and the surface of the cube can be unfolded into a polygon, so the result is nice but not terribly surprising. Of course, the question about the minimal number of pieces remains. –  fedja May 4 '12 at 0:15
Assuming that "pieces" mean "connected polygonal pieces", we can take a 3 by 2 rectangle, cut it into a T-shape and two unit squares and then use the standard "sliding cut" to turn it into a square, giving the total of 6 pieces to cover the unit cube. Can we do better? –  fedja May 4 '12 at 0:34
OK. Posted to soon earlier. The change from the T tetromino to the S tetromino actually allows going down to five pieces instead of six. –  Yoav Kallus May 4 '12 at 4:21
Wow! $\mbox{}$ –  Joseph O'Rourke May 4 '12 at 10:18

Sorry, this is an answer to an other question. (I did not read the question carefully.)

Question: For which $k$, $k$ squares can tile the surface of cube.

Answer: $k=6\cdot(n^2+m^2)$.

Here is a tiling with $k=30$, $n=1$ and $m=2$.

It is obvious if the tiling is vertex-to-vertex.

If the tiling is not vertex-to-vertex, you get a closed geodesic formed by overlaping sides. Then you can shift squares on one side of the geodesic to make the tiling "more vertex-to-vertex". Repeating this operation you can make the tiling to be vertex-to-vertex.

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@Anton: Did you intend $n \ge 1 , m \ge 1$, or, say, $n \ge 1 , m \ge 0$ ? –  Joseph O'Rourke May 3 '12 at 18:03
Yes, $k$ has to be positive; so $n\ge 1$ and $m\ge 0$. –  Anton Petrunin May 3 '12 at 18:33
Awesome ! –  Steven Gubkin May 3 '12 at 19:05
@Anton: Sorry to be slow :-/, but could you describe a partition of the square into 6 pieces that exactly cover the cube? –  Joseph O'Rourke May 3 '12 at 19:21
@Joseph: the partition into faces (the cube has 6 faces). –  Anton Petrunin May 3 '12 at 19:35

No promises that these are optimal, but here are some lower bounds:

With $k=2$, side length $3/8=0.375$ (with one piece flipped over), and with $k=3$, side length $2/5=0.4$:

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