# Equilateral triangle in hyperbolic plane

Show that there is an equilateral triangle with angles $\pi/m$ for any integer $m\ge4$. What is the corresponding result for regular n gon? My attempt: I know that area of triangle in hyperbolic plane is $\pi-(sum \ of \ angles\ of\ triangles)$ . Let y axis be my first line choose second line to be a semicircle with centre right of origin such that it makes an angle $\pi/m$ with y axis .Choose third line to be semicircle with centre on left of origin with angle $\pi/m$ with y axis and the area of triangle by the three lines is $3\pi/m$.

1.Why should the third line intersect second line? 2.How do i write this proof rigourously? I have very little knowledge in this subject so i would prefer an elementary answer.

• Which model of hyperbolic geometry do you use? Using the origin works in the Poincaré disk model but semi circles belong to the Poincare halfplane model don't mix them up – Willemien Jul 13 '17 at 18:19
• I am using upper half plane model – user345777 Jul 13 '17 at 18:51
• by right of origin i mean the semi circle has centre on +ve x axis – user345777 Jul 13 '17 at 18:53
• Not sure if your method will work at all maybe better just calculate the side lengths see en.wikipedia.org/wiki/Hyperbolic_triangle general trigonometry start with the point (0,1) then you can calculate the second point on the y axis and so on – Willemien Jul 13 '17 at 19:17

What is the corresponding result for regular $n$ gon?

A regular $n$-gon is highly symmetric. If you draw all the axes of reflective symmetry, the portion between two such axes is a right-angled triangle. One of its corners is at the center, with an interior angle of $\pi/n$ (because $2n$ of these form the full circle), another is at a vertex of the $n$-gon, with the interior angle being half that of the $n$-gon. The third corner is a right angle at the center of an edge of the $n$-gon.

So if you want interior angles of $\pi/m$ at the corners of a regular $n$-gon, you want to know whether there exists a hyperbolic triangle with angles $\frac\pi{n}, \frac\pi{2m}, \frac\pi2$. The sum of these angles being less than $\pi$ is both necessary and sufficient for the existence of such a triangle.

You can then continue by computing the edge lengths of such a triangle, and then constructing the $n$-gon from that.

for an equilateral triangle with angle $\alpha$ and side $s$ the following relations hold ( see https://en.wikipedia.org/wiki/Hyperbolic_triangle#Trigonometry_of_right_triangles aand a bit of puzzeling)

$$\cos \alpha = \frac {\tanh \frac12 s } {\tanh s }$$ and $$\cosh\frac12 s = \frac {\cos \frac12 \alpha } {\sin \alpha }$$

then with this knowledge you have enough to construct equilateral triangle

for the poincare halfplane

for example point $A = (0,1)$ and $B = \ln(s)$ and then construct the lines under angle $\alpha$ through $A$ and $B$

good luck