A polygon is folded to perfectly wrap a cube, covering all of its surface area with no overlap. Show that the polygon had at least two equal angles.
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$\begingroup$ Crossposted from my post in /r/mathriddles, since the folks over there haven't solved it. $\endgroup$– LopsyFeb 4, 2015 at 13:07
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11$\begingroup$ This feels more like a math problem than a puzzle to me. $\endgroup$– Set Big OFeb 4, 2015 at 14:00
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$\begingroup$ Would it be possible for you to add some context as to where this problem came from, and possibly what you've done to try and solve it? I'd like to send it to Math.SE, but I'm not sure it would fare well there without context. $\endgroup$– user20Feb 10, 2015 at 21:52
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2$\begingroup$ It's a math puzzle that I made up. I suspect it would have done better here if I surrounded it with a story about Alice and Bob needing to deliver a cubical present to a bloodthirsty princess. (If you think so, I'd be happy to add that in, haha.) $\endgroup$– LopsyFeb 11, 2015 at 19:25
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$\begingroup$ Do you know that it is true, or is it an assumption? $\endgroup$– ghosts_in_the_codeFeb 14, 2015 at 16:02
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2 Answers
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I think that the claim is true.
The triangles are, as suggested, chopped and than copied to maintain the coverage that is provided by the classical face structure of a cube. The four cases/points that are mentioned in one of the comments are maintaining of the right angle since the chopped triangles are rotated and the right angle is maintained in the new polygon. This could help in the work but is not yet a proof
Any folding of a cube may be spread to a polygon which is based on the six faces with chopped areas moved from one face to another.
The chopping of areas around the right angles will always result a right angle. The minimal right angle face unfolding contains two right angles which result two right angle in any polygon coverage based on such starting point of creating polygon for coverage. Here is the minimal right angle structure:
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2$\begingroup$ Welcome to Puzzling! Would it be possible for you to edit this answer to explain how your diagram reaches a conclusion about the question? It looks spectacular, but unfortunately, I don't think anyone will understand the point it's making without a little exposition. $\endgroup$– user20Feb 16, 2015 at 9:08
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1$\begingroup$ Don't you have 90 degree angles at C, F1, N, and K? $\endgroup$– xnorFeb 16, 2015 at 9:18
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$\begingroup$ Maybe he means that the angle at each blue point can be assigned a unique value, therefore, no 2 angles will be equal. The shaded part seems to refer to the parts that will be chopped off. $\endgroup$ Feb 16, 2015 at 9:34
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The claim is true.
Proof:
Imagine the system of cuts made into the surface of the cube that allow it to fold open into a flat net. These cuts must have the following properties:
As there are no loops, there must be at least two end points (i.e. vertices of degree 1 in the graph). This means that there are at least two corners of the cube at which a cut ends. When the net is folded out, each of those two corners lead to a polygon angle of 270 degrees (i.e. concave right angle).
- We can consider the cuts as a graph on the cube surface, with straight edges. This is because the resulting net is to be a polygon.
- The cuts must reach each of the 8 cube corners. Any corner without a cut could not be folded flat.
- The graph of cuts does not contain a closed loop. If it did, the cube surface would be bisected, and so not form a single net when unfolded.
- The end point of a cut must be at a corner. If it were not, then the area around the end point would not open up when unfolded, putting the cut's end point in the interior of the flat net. You might as well then undo this superfluous cutting.
The same proof holds not just for a cube, but for any polyhedron with identical vertices.
By the way, polyhedron net shapes can become quite crazy once you allow the cuts to go across the polyhedron faces rather than just following its edges. For example, below is a picture of weird cube net. With a few tweaks its identical angles can be removed, but whatever you do you cannot get rid of the two 270 degree ones.
answered Jul 12, 2017 at 16:10
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$\begingroup$ Nice proof. Looks to be along the same lines as the one from Reddit (linked in comments on the question). Hopefully Lopsy is still around to accept this answer. $\endgroup$ Jul 13, 2017 at 12:35
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$\begingroup$ @Randal'Thor: I hadn't noticed that it was an old question that had been bumped for some reason, and took Lopsy's comment that it hadn't been solved on Reddit at face value. I think that if I had noticed, I would have checked the links and not bothered adding a my answer. $\endgroup$ Jul 13, 2017 at 13:51