The sum of angles of a hyperbolic triangle is $\alpha+\beta+\gamma=\pi-A(\Delta)$ with the area $A(\Delta)$ of the triangle. Thus, if we construct a series of hyperbolic triangles where the area converges to $0$, the angle sum should approach $\pi$. I tried doing this with the following explicit construction of a triangle $z_1,z_2,z_3$ in the half-plane model:
Given two circles of radius $1$ and center $\pm 1/2$, the intersection is in $$z_1:=\frac 32 \mathrm i.$$ Then, we take the other points to be $$z_2 := -\frac 12 + \mathrm e^{\mathrm i\varphi},\quad z_3 := \frac 12 + \mathrm e^{\mathrm i(\pi-\varphi)},$$
i.e. arbitrary points on the two circles. Both points approach $z_1$ for $\varphi\to \pi/3$.
Because $z_2,z_3\to z_1$ for $\varphi\to \pi/3$, the area of the triangle should approach $0$ and the sum of angles should approach $\pi$.
This is what I thought; however, a direct calculation of the sum of angles leads to a different result.
The angle in $z_1$ is the angle between the radii of the two circles from the beginning which intersect in $z_1$, i.e. between $z_1+1/2$ and $z_1-1/2$, so
$$\alpha = \arccos\left(\frac{\Big\langle\begin{pmatrix}1/2 \\ \sqrt 3/2\end{pmatrix}, \begin{pmatrix}-1/2\\ \sqrt 3/2\end{pmatrix}\Big\rangle}{1\cdot 1}\right) = \arccos \frac 12 = \frac \pi3.$$
The angle in $z_2$ is the angle between the radii of the initial circle and the dashed circle in $z_2$, i.e. between $z_2+1/2$ and $z_2$, so
$$\beta = \arccos\left(\frac{\Big\langle \begin{pmatrix}\operatorname{Re}z_2+1/2 \\ \operatorname{Im}z_2\end{pmatrix}, \begin{pmatrix}\operatorname{Re}z_2 \\ \operatorname{Im}z_2\end{pmatrix}\Big\rangle}{1\cdot |z_2|}\right)=\arccos\left(\frac{|z_2|^2+(\cos\varphi-1/2)/2}{|z_2|}\right),$$ where $|z_2|^2=(\cos\varphi-1/2)^2+\sin^2\varphi=5/4-\cos\varphi$, so $$\beta=\arccos\left(\frac{1-(\cos\varphi)/2}{\sqrt{5/4-\cos\varphi}}\right)\to\arccos\frac{\sqrt{3}}{2}=\frac \pi6$$ for $\varphi\to\pi/3$.
The third angle $\gamma$ is equal to $\beta$ due to symmetry.
So, we get $\alpha+\beta+\gamma\to 2\pi/3$ for $\varphi\to\pi/3$, but $A(\Delta)\to 0$.
Where is the mistake; how does this make sense? Thanks!
Best Answer
The interior angle at $z_1$ is actually $2\pi/3$. You computed the exterior angle, not the interior one.