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My task is to wrap a unit cube with the smallest square sheet of paper possible. The paper is assumed to be infinitely thin of course and no cutting or stretching is permitted.

I must be able to justify any claims I make.

This is what I have so far:

Let $A$ denote this smallest area I seek. (How do I know there is a smallest area? Wouldn't it be the greatest lower bound of a monotonically decreasing sequence of $A$'s with lower bound 6?)

A well-known fact from geometry says that a cube has 11 nets. Since none of the nets are square, my sheet of paper can't be folded like any of them, so $A>6$.

A square $2\sqrt{2}$ on a side can wrap a unit cube. Here's how: make four folds each perpendicular to a diagonal and 3/2 from the paper's corner. Place the cube in the resulting square created by the folds. The corners of the paper will meet at the center of the side on top. Hence, $A\le (2\sqrt{2})^2=8$.

I am stuck at $6<A\le 8$. I can't find anything smaller nor can I prove the square of area 8 is the smallest.

Any suggestions will be much appreciated.

Update: $A=8$. This was proved by Michael L. Catalano-Johnson, Daniel Loeb and John Beebee in "Problem 10716: A cubical gift," American Mathematical Monthly, volume 108, number 1, January 2001, pages 81-82 (posed in volume 106, 1999, page 167).

So what do I do now as far as accepting an answer? Would it be frowned upon if I summarized their solution down below? (That way, I might get someone to comment on it, as I don't completely follow.)

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Thinking aloud here... each of the 8 corners must be covered at least once by the sheet. And for each point on the sheet that covers a corner, a certain minimal amount of paper must be wasted on folds, and this amount seems to be related to how much of the adjoining edges the local cover reaches. Perhaps a lower bound could be derived from the fact that all 12 edges must be fully covered? But it's probably not that easy -- for convex (but not necessarily square) sheets that will cover the cube there's an easy upper bound of 7.5 (which can be lowered to $8-\sqrt{1/2}$ with a bit of care). –  Henning Makholm Oct 28 '11 at 13:22
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Look at: mathpuzzle.com/SerhiyWrapping.html –  Joseph Malkevitch Oct 28 '11 at 15:55
    
It's perhaps of note that it takes 2 points to enclose a unit segment, and 4 unit segments to enclose a unit square. I would not be entirely surprised if the trend of $2^n$ continued. –  A Walker Oct 28 '11 at 19:36
    
@JosephMalkevitch From the site you linked to, I was able to chase down a few references, one of which pointed to a full solution. That reference is: Michael L. Catalano-Johnson and Daniel Loeb, "Problem 10716: A cubical gift," American Mathematical Monthly, volume 108, number 1, January 2001, pages 81-82 (posed in volume 106, 1999, page 167). Thanks! –  sasha Oct 28 '11 at 19:53
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@sasha: No, it would not be frowned upon at all! Look at this blog post from Jeff Atwood himself. –  Zev Chonoles Oct 28 '11 at 22:31

2 Answers 2

up vote 3 down vote accepted

I am going to summarize the trio's solution here since it's essentially an answer to my question, but I'm hoping that the first paragraph in particular might be clarified by one of you as I don't completely follow. The problem, titled "A Cubical Gift" [10716], was posed in the 1999 volume of the Monthly and phrased differently ("What is the largest cubical present that can be completely wrapped (without cutting) by a unit square of wrapping paper?") Clearly, an answer to this question will solve mine and vice-versa.

Two arbitrary points in the square are chosen and their distance is considered before and after wrapping the cube with the square. They argue that the surface distance between the two points after wrapping is no larger than what it was before wrapping, since the paper was neither cut nor streched. [I'm not convinced. Perhaps the surface distance notion is throwing me off. Is it the distance along the paper? Wouldn't that then be an obvious statement, a tautology even? Otherwise, are we talking about the distance along the cube? Isn't there a lot of choices to call the distance in that case? And how does this paragraph relate to the next?]

Consider an arbitrary point on the cube. There exists another at least twice an edge length away [surface distance, again]. This implies that for any point on the square, another point can be found at distance at least twice an edge length. So, this is also true for the center of the square, implying that the diagonal is at least 4 times the length of an edge. Therefore, the side length of the paper square is at least $2\sqrt{2}$, giving the area of 8. Finally they show that the inequality is tight by demonstrating the folding pattern I described in the OP.

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By surface distance between two points they mean two things depending on whether the two points are on the sheet or on the cube. On the sheet it is the usual distance between two points. On the cube it is the length of the shortest path along the surface of the cube. If the straight line segment from point $A$ to $B$ on the sheet of paper has length $L$, then, after wrapping, tracing the image of that line on the cube gives a path of length $L$ (no stretching). That path may not be the shortest route connecting the image points $A'$ and $B'$ on the cube, hence you only get inequality. –  Jyrki Lahtonen Oct 29 '11 at 6:27

Consider three faces surrounding a corner of the cube. If we suppose that the edge of the square piece of paper is nowhere cutting across these three faces, then in the folding process we inevitably `double cover' near the corner. This waste of paper will, by concavity, amount to at least $1/2$ of a unit of area. If we suppose that only one face, $F$, has a paper edge crossing it, orient $F$ to the top. Then the four other faces surrounding the bottom face each satisfy the above criterion, and we waste at least $2$ units of paper (corresponding to the four bottom corners). This implies that one must use at least 8 units of paper to cover the cube, if one face is crossed by a paper edge.

If no faces are disturbed in this way, then a similar bound holds. Either your bound is optimal or there exists a more efficient one in which between 2 and 6 faces meet the paper's edge.

Hope this helps as a technique. (It seems likely that 8 is optimal.)

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