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Given a square piece of paper (say 1x1), what is the largest cube net you can cut out of it?

('largest' meaning with maximal volume; 'net' meaning it's possible to fold into a cube (somehow))


This puzzle was stolen from 'The bedside book of geometry' (Mike Askew and Sheila Ebbutt) which I happened to find lying around in the house

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  • $\begingroup$ First thing that came to mind was actually to make a cube with just a ribs. Too bad that would mean the answer is trivial. $\endgroup$ – Dennis Jaheruddin Apr 23 '18 at 13:11
  • $\begingroup$ @Therandomguy Please post answers as answers, not comments. $\endgroup$ – David Richerby Apr 23 '18 at 16:11
  • $\begingroup$ Is there a requirement that there not be any overlapping of excess material? If not, I think the answer would be arbitrarily close to a cube whose sides have a total area equal to that of a square [absent such a requirement, an arbitrarily slender rectangle can be cut and folded to yield a connecting member of arbitrary length, so one could use just about any dissection which converts a square into a cube net, but shrink the pieces just enough to add connecting members.] $\endgroup$ – supercat Apr 23 '18 at 16:16
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The largest cube net I can cut out of a 1x1 square paper has a volume of

$ \frac{\sqrt{2}}{32} \approx 0.044 $

Using the following cutout:

The black area is exactly $\frac{6}{8}$ of the total area (The white triangles are half the area of one face), which means each face has an area of $\frac{1}{8}$, giving a volume of $(\frac{1}{\sqrt{8}})^3 = \frac{\sqrt{2}}{32}$. Alternatively we can notice that the length of the diagonal of a face of the cube is $\frac{1}{2}$.

X-shaped cutout of a cube

Comparison of volumes from different methods:

$\begin{array}{l|r|r|r} \text{Method} & \text{Volume} & \text{Approx.} & \text{Percentage} & \text{Wasted paper} \\ \hline \text{Theoretical maximum} & \frac{\sqrt{6}}{36} & 0.068 & 100.0 & 0 \\ \text{wl} & \frac{\sqrt{2}}{32} & 0.044 & 65.0 & \frac{1}{4} \\ \text{CiaPan} & \frac{\sqrt{10}}{100} & 0.032 & 46.5 & \frac{2}{5} \\ \text{Default net along diagonal} & \frac{2\sqrt{2}}{125} & 0.023 & 33.3 & \frac{13}{25} \\ \text{Default net along side} & \frac{1}{64} & 0.016 & 23.0 & \frac{5}{8} \\ \end{array}$

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  • $\begingroup$ Nice one, particularly how you form one face from the corners. Could you also provide the working how you arrived at the final answer? $\endgroup$ – Phylyp Apr 23 '18 at 9:46
  • $\begingroup$ @Phylyp You mean the volume or how I came up with this cutout? $\endgroup$ – w l Apr 23 '18 at 11:28
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    $\begingroup$ The comparison is nice. Any proof that this solution cannot be improved upon? $\endgroup$ – Shufflepants Apr 23 '18 at 19:12
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    $\begingroup$ I think it can be proven from the fact that the longest straight line available is the diagonal of the square paper, and the longest straight line you can draw on a cube is its circumference passing through 4 faces. This cutout maps the former to the latter.. $\endgroup$ – Amit Naidu Apr 24 '18 at 5:38
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    $\begingroup$ @wl But that line, if shifted sligthtly to a parallel plane, appears not 'straight': it turns by 45° when crossing the cube's edge. $\endgroup$ – CiaPan Apr 24 '18 at 9:20
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This:

enter image description here

fits a net of a cube with an edge length $a$ into a square with a side of

$$a\,\sqrt{10}$$

so for a square of side length equal $1$ we get a cube with edge length

$$a=\frac 1{\sqrt{10}} \approx 0.316$$

and volume

$$V=a^3 = \frac 1{10\sqrt{10}} \approx 0.0316$$

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  • $\begingroup$ The previous answer has a greater volume though and is way more simple $\endgroup$ – Gintas K Apr 23 '18 at 11:33
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    $\begingroup$ @GintasK Sure, but it's still an interesting approach. $\endgroup$ – maxathousand Apr 23 '18 at 13:15
  • $\begingroup$ @GintasK Yes, I know. But when I started solving, the previous answer has not been posted yet. Why not post what I've done, just because someone else was faster? Anyway, I hope this site doesn't have any rule against publishing answers which are worse than those already posted... $\endgroup$ – CiaPan Apr 24 '18 at 8:03
  • $\begingroup$ This question reminded me a similar problem, which I met years ago in some maths newsgroup of Usenet: fit an unfolded net of a cube into an ISO 216 sheet, that is into a 1:sqrt(2) rectangle. The solution above is somewhat similar to that old answer I gave there (which possibly also was not optimal). $\endgroup$ – CiaPan Apr 24 '18 at 8:04

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