Construct a quadrilateral in a Poincare half plane

[Robb]

Homework Statement

Construct a quadrilateral in the poincare half-plane model, such that the sum of all its angles is less than 60 degrees.

Relevant Equations

n/a

I created the attached file in geogebra but I don't know how to come up with less than 60 degrees. There always seems to be at least one angle that is large, no matter how I manipulate the curves. Quadrilateral is points US(1)V(1)W. The attached does not include angle measures but I assume the idea is clear. Any help would be much appreciated!

 

[Antarres]

To be honest, I'm not really versed in working on this geometry, but from what I see, geodesics(a.k.a. 'straight lines') in Poincare half-plane model, are circles perpendicular to the $x$-axis, as well as vertical lines, perpendicular to $x$-axis. I haven't checked this, just googled it, but if I'm right about that, would a quadrilateral that would be produced if you put a vertical line through $S_1$ on your picture eliminate the big angle you're pointing out?

I haven't calculated the angles in this metric, but maybe that helps. Report if it makes a difference.

P.S. I hope I'm not going too far here with the solution, I'm just addressing the problem you're talking about, though I can't see the angles measured in this metric, so I wouldn't know if it solves anything. Maybe a better explanation of your attempts helps in that case.

 

[Robb]

To be honest, I'm not really versed in working on this geometry, but from what I see, geodesics(a.k.a. 'straight lines') in Poincare half-plane model, are circles perpendicular to the $x$-axis, as well as vertical lines, perpendicular to $x$-axis. I haven't checked this, just googled it, but if I'm right about that, would a quadrilateral that would be produced if you put a vertical line through $S_1$ on your picture eliminate the big angle you're pointing out?

I haven't calculated the angles in this metric, but maybe that helps. Report if it makes a difference.

P.S. I hope I'm not going too far here with the solution, I'm just addressing the problem you're talking about, though I can't see the angles measured in this metric, so I wouldn't know if it solves anything. Maybe a better explanation of your attempts helps in that case.

Attached, I have added the perpendicular line but still the same issue.

 

[Antarres]

I don't see any difference between the attachments you have posted.

 

[Robb]

I don't see any difference between the attachments you have posted.

Sorry, must have sent the wrong file. See attached.

 

[Antarres]

Okay, I think there can be this type of solution. In hyperbolic geometry, we define ideal triangles as triangles which have sum of angles equal to zero, that is, their vertices are at the boundary of hyperbolic space. In Poincare upper half-plane model, the points on the real line are ideal points, as well as points when $y \rightarrow \infty$. So we can construct this type of ideal triangle in this model:

This triangle has all angles equal to zero, and putting two of those together, we get the so called ideal quadrilateral who's angles are also all equal to zero.
So the idea when constructing smaller and smaller angles, is that you want the vertices to be closer to the $x$-axis. In case when all are on $x$-axis(or in vertical infinity), you get zero angles. This way, you can even adjust the sum of angles however you like, since you can slowly deform the ideal triangle(or quadrilateral) so that the sum of angles becomes greater than zero and as big as you want it to be(of course, always smaller than $2\pi$ radians). Hope this helps some more.

 

[Robb]

This is perfect! Thanks so much. Makes sense. I didn't consider sending y to infinity. Good day, my friend!

 

[Antarres]

You're welcome!

 

[Robb]

You're welcome!

FYI, attached is another option.

 

[Antarres]

FYI, attached is another option.

Yeah, all vertices can be on the $x$ axis, as well. Good work!