# Simple proof of existence of hyperbolic triangles

I've studied the hyperbolic plane by analytically building up the hyperboloid model, the Klein—Beltrami disc, the Poincaré disc, and the half-plane model from scratch. Now I'd like to prove that, given $\alpha,\beta,\gamma>0$ with $\alpha+\beta+\gamma<\pi$, there exists a hyperbolic triangle with angles $\alpha$, $\beta$ and $\gamma$.

Is there an easy proof of this fact? I mean a proof that only starts from the basic Riemannian structure of the models: the knowledge what the geodesics are, an expression for the distance, the fact that some models are conformal. (Of course, I have found proofs in literature, but they tend to refer a lot to other results, requiring e.g. an extensive theory of trigonometry.)

In the disc model take two lines $L_1,L_2$ emanating from the center $0$, and forming an angle $\alpha$ berween them. Now, consider the point $x_t\in L_1$ ad distance $t$ from $0$ and the line $L_3(t)$ emanating from $x_t$ and forming an angle $\beta$ with $L_1$. Let $y_t=L_3(t)\cap L_2$. Now, $y_t$ can be a single point or the empty set. If $y_t$ is a point, you get a triangle with angles $\alpha,\beta,\gamma(t)$.

You can check that $\gamma(t)$ varies continuously on $t$ and that

1) as $t\to 0$, $y_t$ is a point and the sum $\alpha+\beta+\gamma(t)\to\pi$

2) There is the first time $T$ such that $L_3(t)\cap L_2$ is empty. As $t\to T$ you have $\gamma(t)\to 0$.

Given $\gamma<\pi-\alpha-\beta$, by means of the intermediate value theorem you see that there is $t$ so that $\gamma(t)=\gamma$

• Nice idea, but the intermediate value theorem cannot be used to pinpoint a point in the $(t,s)$-plane. Try fixing the angle $\beta$ between $L$ and the moving leg. – Christian Blatter Jul 28 '14 at 10:58
• sure, "by means of" was maybe to vague. Fixing $\beta$ and move, as you suggest, definitely work. – user126154 Jul 28 '14 at 11:03
• @ChristianBlatter I Edited the answer according your comment – user126154 Jul 28 '14 at 11:11
• You're swapping $L_1$ and $L_2$ in the beginning, and there are multiple such lines $L_3$, but I see how this gives an easy solution I was looking for. – Socci Jul 28 '14 at 13:21
• @Socci Corrected the swap of $L_1$ and $L_2$. As for the lines $L_3$ yes, you're right, a priori you have two of them, but if you make a picture it is clear which one to choose. May be I could edit by introducing oriented angles so that the construction wil be unique. – user126154 Jul 28 '14 at 14:53

Yes you can

Assuming the curvature to be $\sqrt{-1 }$ the length of the side c is

$$c = \operatorname{arcosh} \left( \frac{\cos \alpha \cos \beta + \cos \gamma }{\sin \alpha \sin \beta} \right)$$

(found the formula in "the foundations of geometry and the non euclidean plane" by Martin corollary 32.16 )

so you can even construct your triangle :)

edit : corection of formula

The lenght of the sides are:

$$a = \operatorname{arcosh} \left( \frac{\cos \beta \cos \gamma + \cos \alpha }{\sin \beta \sin \gamma} \right)$$

$$b = \operatorname{arcosh} \left( \frac{\cos \alpha \cos \gamma + \cos \beta }{\sin \alpha \sin \gamma} \right)$$$$c = \operatorname{arcosh} \left( \frac{\cos \alpha \cos \beta + \cos \gamma }{\sin \alpha \sin \beta} \right)$$

then you can make the triangle (all sides and angles are known, what more do you need?)

• If you meant area-hyperbolic-cosine, I'd suggest using \operatorname{arcosh} for the function name. This forces a word to be set with upright font. You might also want to replace braces with \left\{ and \right\} version to have them automagically adjusted to their contents size. – CiaPan Jul 28 '14 at 11:42
• @CiaPan done thanks – Willemien Jul 28 '14 at 11:47
• @Willemien Sorry, I don't really follow. – Socci Jul 28 '14 at 13:07
• i corrected the forma, with these details everything of the triangle is known, ,then you can construct the triangle – Willemien Jul 28 '14 at 18:30