Vertices of hyperbolic triangle with given angles

Question

This is probably a well-known problem in hyperbolic geometry, but here goes anyway. In the Poincar'e upper-half plane model, I am given three angles $\alpha$, $\beta$, and $\gamma$ with $\alpha+\beta+\gamma<\pi$. I would like to find vertices $A$, $B$, and $C$ of a hyperbolic triangle so that the respective angles are $\alpha$, $\beta$, and $\gamma$. Of course I can always solve equations for this, but if possible my requirement is that the coordinate of the vertices be as "simple" as possible, more precisely in a number field of small degree when the angles are rational multiples of $\pi$. I know how to do this when one angle is zero (CM points), but in general it seems the degree is 4. Is there some "canonical" way to do this ?

Answer 1

The simplest way I know is to use the second hyperbolic law of cosines to find the lengths of the sides of the triangle, which then reduces the problem to taking two square roots plus some rational manipulations.

The point is this: If the angles of the triangle are $\alpha_i$ ($i=1,2,3)$ and the length of the side opposite the $i$-th vertex is $c_i$, then the second hyperbolic law of cosines says that $$ C_i = \cosh(c_i) = \frac{\cos\alpha_i + \cos\alpha_j\cos\alpha_k}{\sin\alpha_j\sin\alpha_k}, $$ for $i$, $j$, and $k$ distinct.

Now, using the hyperboloid model, where $H = \{(x,y,z)\in\mathbb{R}^3\ |\ x^2+y^2-z^2 = -1\ \text{and}\ z>0\ \}$ and the Minkowski inner product, it follows that the vertices of the triangle are three points $P_i\in H$ where $P_i\cdot P_j = -C_k$. By a symmetry of the model, you can assume that, say, $P_1 = (0,0,1)$ and that $P_2 = (0,a,b)$ (where $a$ and $b$ are positive) and $P_3=(p,q,r)$. Then $b = -P_1\cdot P_2 = C_3$ and $r = -P_1\cdot P_3 = C_2$, implying that $a = \sinh(c_3) = \sqrt(b^2-1)$.

Now, $-C_1 = P_2\cdot P_3 = aq-br$, so you can solve for $q$, as $a,b,r, C_1$ are known and $a>0$. Finally, since $p^2 = r^2-1-q^2$, you can solve for $p$ by taking a single square root. It follows that by adjoining at most two square roots to the field that contains the $C_i$, you can explicitly find $P_1,P_2,P_3\in H$ in that field extension of degree at most $4$.

Finally, if you want to convert these $P_i$ to points in the upper half-plane $U$, you can use the map from $H$ to $U$ given by sending $(u,v,w)\in H$ to $$ \left(\frac{u}{1+w},\frac{v}{1+w}\right)\in U, $$ which is a rational map, so the coordinates of the images of the vertices $P_i$ will lie in the same field extension of degree at most $4$ of the field that contains $C_1$, $C_2$, and $C_3$.

Answer 2

We can work in the disc model consisting of all complex numbers of norm $<1$. We denote by $\alpha, \beta,\gamma$ unit vectors making the correct angles with the positive real axis (we have thus $\alpha\beta\gamma<1$). We put the vertex with angle $\alpha$ at $0$ with adjacent sides included in strictly positive real numbers and in $\alpha\mathbb R_{\geq 0}$. The two remaining vertices are thus given by $t\alpha$ (with angle $\beta$) and $s$ for suitable $r,s\in(0,1)$. The center of the circle defining the side containing the vertices $t\alpha$ and $s$ is given by $c=\alpha t-u i\alpha\beta= s+u i/\gamma$ with $u$ a strictly positive real number. This gives a real linear system with $2$ equations in $s,t,u$. We need moreover orthogonality of the circle of radius $u$ centered at $c$ with the unit circle. This gives the equation $\vert c\vert^2=1+u^2$ which is quadratic. I guess that one solution is negative and the other one is the correct solution.

In order to get the the upper-halfplane, a suitable involution involving $i$ is needed (this and the occurence of $i$ in the above formulae together with the quadratic equation explains probably that the final degree is $4$).