Is there a surface on which a hexagon can have all right angles? 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?


You would need a surface of negative curvature.

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).
