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.
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.