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So I was watching a video that features astronomer and topologist Cliff Stoll talking about how figures that aren't quadrilaterals can have all their angles equal 90 degrees on different surfaces. For example, on a sphere, you can create a triangle that has all of its angles equal $90^\circ$. On a pseudosphere, you can create a pentagon that has all of its angles equal $90^\circ$. Now, here's my question.

Is there a surface where a hexagon with this property is possible?

amWhy
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Xavier Stanton
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    On hyerbolic surfaces you can have a triangle where the sum of the angles is 0. And hexagons with 6 right angles. – Doug M Nov 30 '18 at 01:38
  • you can find many questions on this site about the hyperbolic plane, usually in one of the popular models: disc model, upper half-plane model. The bad news is that we cannot fit the whole thing into a pseudosphere... – Will Jagy Nov 30 '18 at 01:39
  • Related: https://math.stackexchange.com/q/691894/90543 – jgon Nov 30 '18 at 01:39
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    @DougM I think it is worth to note that the existence of a triangle where sum of angles is 0 depends on the definition of the triangle -- it is not clear whether ideal triangles are allowed, since their vertices are in infinity. If we require all the vertices to be in the hyperbolic plane, then the angles have to be greater than 0, and can be as small as we want. – Zeno Rogue Nov 30 '18 at 11:40
  • Searching "hyperbolic crochet" on Google Images shows lots of pictures of the waviness that is the hyperbolic plane (or, at least, small sections of it) embedded in three-space. (It should be noted that the areas of circles on the hyperbolic plane grow exponentially as you increase the radius—unlike the flat Euclidean plane where they grow quadratically—meaning that, assuming the yarn has nonzero thickness, you'll quickly run out of space as you try to include larger and larger sections of the hyperbolic plane.) – Akiva Weinberger Nov 30 '18 at 12:05
  • Here's one where the radius is particularly large compared to the curvature. And here's a more sensibly sized one, with the parallel postulate illustrated. (Those are all straight lines, since you can fold along them.) – Akiva Weinberger Nov 30 '18 at 12:17
  • I realize I'm writing a lot down here… Turns out Taimina made a TedX talk about this so you should just watch that: https://youtu.be/w1TBZhd-sN0 – Akiva Weinberger Nov 30 '18 at 12:53

3 Answers3

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You would need a surface of negative curvature.

illustration of a hexagon in the hyperbolic plane

It is best to use a hyperbolic plane for this, where you can easily fit any regular n-gon with given angles as long as the sum of its external angles is greater than 360 degrees. The problem is that the hyperbolic plane does not fit in Euclidean space.

The pseudosphere is a small fragment of the hyperbolic plane. You can draw a right-angled hexagon on the pseudosphere only if you allow it to wrap over itself. (Edit: actually I am not completely sure about this; see here, you get a pseudosphere by cutting the part covered with white dots; it appears that a hexagon is slightly larger than the area covered by the pseudosphere, but I am not sure. Should be possible to prove.)

You can also draw it on a Dini's surface -- that is basically an unrolled pseudosphere where you have several layers, and thus you avoid the intersection problem. But it would be hard to see anything because it is rolled very tightly. See here.

Less smooth, but probably the best way would be to use something similar to a hyperbolic crochet. See our computer simulation (arrow keys to rotate).

psmears
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Zeno Rogue
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  • Other answers provides other (possibly simpler) planes, but they don't have constant curvature. – user202729 Nov 30 '18 at 11:00
  • It's been such a long time since I last played your game! I can't check now because I'm on mobile, but did you ever end up adding the azimuthal equidistant projection? – Akiva Weinberger Nov 30 '18 at 13:17
  • On a related note, I just rewatched this video. The ending is so shocking! – Akiva Weinberger Nov 30 '18 at 13:21
  • Sure! I have added azimuthal equidistant projection and some other cool projections too. They are listed here. – Zeno Rogue Nov 30 '18 at 13:23
  • Ah, nice! (Oh my god it looks so weird) – Akiva Weinberger Nov 30 '18 at 13:33
  • This is a cool concept. The thing is though, it seems like a pseudosphere tends to infinity. This is because a pseudosphere is defined as having constant negative Gaussian curvature. But if the pseudosphere was not tending to infinity, there would be a small part of the shape with positive Gaussian curvature, which contradicts this characteristic. So, how do you make the pseudosphere wrap over itself if this is the case(of it tending to infinity in one direction)? – Xavier Stanton Nov 30 '18 at 20:38
  • I do not understand what you mean. How is wrapping over related to tending to infinity? We do not actually use the end which tends to infinity (the hexagon is drawn in the "wide" part), so we can simply cut it away. – Zeno Rogue Nov 30 '18 at 21:11
  • By wrapping over itself, I meant like in the Dini's surface, but with the pseudosphere touching itself without leaving any space. – Zeno Rogue Nov 30 '18 at 21:13
  • Okay, sorry for the misunderstanding. – Xavier Stanton Dec 01 '18 at 23:27
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I'm afraid the constraints in the question as stated might be lower than intended, because it isn't difficult to fit a surface to most of 6-segment right-angled closed lines in space in a way that all angles lie in that surface.

Taking in account that the OP hasn't even stated that all right angles must turn to the same side, there is an even simpler solution: a cylinder. We just need to draw a 6-segment zigzag with right angles across a piece of paper and fold the paper in a cylinder to close the line.

If the fact the cylinder has has a hole is a problem, we can close it with a semi-sphere - not included in the photos because I didn't have at hand a ball of a suitable size.

Addition of a solution with all angles to the same side (no zigzag):

I beg your pardon for my poor drawing skills.

Please take two contiguous faces of a cube, make a circuit using all edges except for the common one, and attach an small square to each of the six vertexes, in the plane of the circuit.

enter image description here

Now put one band along each edge the circuit, joining the squares. Put it in a way that all turns at the vertexes keep to the same side. In the next picture I added normal vectors to the squares, twists and turns for clarification:

enter image description here

And since there are no complete twists, you can close the hole in the band to form a disc:

enter image description here

Pere
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    This has to be a hexagon, not a zigzag line with 6 segments. A hexagon has an outside and an inside, and it cannot have a hole in the inside. Also the angles have to be 90 degrees on the INSIDE (otherwise they are actually 270 degrees). It is easily possible to make a hexagon with 90+90+90+90+90+270 degrees even on a plane, that would be an L shape. You need a negatively curved surface to make it 90+90+90+90+90+90, it will not be possible on a cylinder or a semi-sphere. – Zeno Rogue Nov 30 '18 at 11:09
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    @ZenoRogue - As I said in the answer, if you need an inside and an outside, just cover the hole of the cylinder with half a sphere. And an L-shaped hexagon is still a hexagon. Furthermore, there are other solutions with all angles to the same side: you just need to fix a twisted band to the edges of a cube. Of course, you would be probably right if the question asked for a regular hexagon in an constant curvature surface, but that's not in the question - even if the OP had it in mind, which is possible. – Pere Nov 30 '18 at 11:20
  • How would you make a solution with all angles on the same side by attaching a twisted band? If you do your cylinder solution on a Klein bottle (or a torus), then it has no outside, so it is not a hexagon either. If you make two right-angled triangles and connect them with a handle, this is not a hexagon either, because it is not homeomorphic to a disk. – Zeno Rogue Nov 30 '18 at 11:31
  • @ZenoRogue - For an hexagon it would be homoeomorphic, because we just need to twist it 90º right and left in succession, two times. – Pere Nov 30 '18 at 11:54
  • I do not understand what you mean... – Zeno Rogue Nov 30 '18 at 11:58
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    @ZenoRogue - I added an explanation and some drawings. I hope it helps. – Pere Nov 30 '18 at 12:29
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    Thanks! I understand now, and it works indeed -- it is easier to see when we flip the odd corner to the right. By Gauss-Bonnet theorem the curvature of this surface must be negative somewhere. – Zeno Rogue Nov 30 '18 at 13:31
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    At least, the twisted bands have negative curvature in my rendition. Other parts have positive curvature or no curvature. – Pere Nov 30 '18 at 19:50
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Take a cube and orient it so you're looking down a body diagonal. The edges that appear to be at the boundary of this view actually form a nonplanar object with six congruent edges, right angles wherever two adjacent edges meet, and $D_{3d}$ point group symmetry.

Oscar Lanzi
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