I would like to construct a regular right angled hexagon in a klein model.
I'm having a hard time understanding why this method works, here is what my professor did in class. Any additional comments will be greatly appreciated. Thanks in advance!
First in a circle O, pick any six points, $A,B,C,D,E,F$, connect $AB$, $CD$, and $EF$.
Then, construct line $a, b, c, d, e, f$ where $a,b$ intersects, $c,d$ intersects, and $e,f$ intersects, call them points $AA, BB,$ and $CC$.
Lastly, connect $AA, BB, CC$ to construct lines $\alpha, \beta,$ and $\gamma$, the hexagon inscribed in the circle hexagon ($\theta1 - \theta6$) all have right angles therefore is a regular hexagon.
You will not make it that way, what your professor showed you is how to make an equiangular hexagon (all angles are equal in measure) in a Beltrami Klein disk model , you can transform that into an equiangular hexagon in a Poincare disk model, but that is a later worry.
The first step is to get a regular hexagon in a Beltrami Klein disk model
devide the circle in 12 equal arcs and number the dividing points. (for nice looks don't start at 12 0'clock but start halfway between 11 and 12)
draw segments between the points 1 and 4, 2 and 11, 3 and 6, 5 and 8, 7 and 10, 9 and 12
You now have a regular hexagon in a Beltrami Klein disk model. ( you can check that all angles are right hyperbolic angles by the method given by your professor)
Then for every segment:
draw a line tangent to the circle for each of the endpoints of the segment.
draw a circle from the intersection of these lines , trough the endpoints of the segment (this circle should go to each endpoint and be orthogonal to the circle)
Do this for every of the segments 1-4, 2-11, 3-6, 5-8, 7-10, 9-12
Done !