# Constructing a regular right angled hyperbolic hexagon

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. • "... the circle ($\theta_1$-$\theta_6$) ..." : From my scratch-work, there's no guarantee that those six points lie on a common circle. – Blue Feb 27 '14 at 1:46
• @blue, sorry i meant it being a hexagon. I'll fix that. – PandaMan Feb 27 '14 at 6:16
• What does "the hexagon inscribed in the hexagon" mean? The hyperbolic hexagon with the same vertices as the Euclidean one you constructed? Again from my scratch-work, the hyperbolic lines through those vertices don't appear to meet at right angles. – Blue Feb 27 '14 at 6:31
• Oh, wait ... You must be using the Klein-Beltrami Disk, not the Poincaré. Each $\theta_i$ is a right angle simply by the interpretation of perpendicularity (for non-diameters) in the KB model. – Blue Feb 27 '14 at 6:48

## 1 Answer

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 !