# Arbitrary hyperbolic triangle

I have a following exercise which I am struggling to believe is possible.

Let $T$ be an arbitrary hyperbolic triangle in the Poincaré Disc $D$, with vertices $a,b,c$ $\in$ $D$, and sides $[a,b], [a,c], [b,c]$. Let $p_a ∈ [b, c]$, $p_b ∈ [a, c]$ and $p_c ∈ [a, b]$ be such that

$d_\mathbb D(a, p_b) = d_\mathbb D(a, p_c)$, $d_\mathbb D(b, p_a) = d_\mathbb D(b, p_c)$ and $d_\mathbb D(c, p_a) = d_\mathbb D(c, p_b)$

(a) Prove that for any points $q$ $\in$ $[a,p_c]$ and $r$ $\in$ $[a,p_b]$ with $d_D(a,q) = d_D(a,r)$, there exists a constant $\delta$ $\in$ $[0,\infty)$ such that $d_D(q,r)\leqslant \delta$.

(b) Conclude that any side of a hyperbolic triangle in the hyperbolic plane lies in the closed $\delta$-neighbourhood of the union of the other two sides.

I originally thought that choosing $\delta$ to be the maximum of $d_D(b,p_c)$ and $d_D(c,p_a)$ would work but then realised that I have assumed the point closest to $p_c$ on one of the other sides is $p_b$, when it could lie on $[b,c]$.

Do you have to find a $\delta$ that depends on just $b$, and $c$? Or any $\delta$ is fine providing it doesn't depend on $a,r$ and $q$? Or does part (a) mean that we have to show $d_D(q,r)$ is finite?

Thanks

• How about $\delta = d_{\Bbb D}(a, p_b) + d_{\Bbb D}(a, p_c)$? That is the $\delta$ that works in Euclidean geometry, and the hyperbolic distance should still satisfy the triangle inequality. – Paul Sinclair Dec 28 '17 at 23:47
• Please don't edit your question in such a way that it changes to a completely different (and in this case rather poorly worded) question. I've rolled back the edit; please don't repeat it. If you want to ask a different question, ask it as a new question. – MvG Jan 2 '18 at 20:00

Sometimes a good illustration can help a lot. Start by looking at the initial definitions in a Euclidean world. Drawing corners $a,b,c$ for a generic triangle, the points $p_a,p_b,p_c$ are already uniquely defined. Where are they? Here:
2. Would you happen to have any good idea for a suitable bound $\delta$ in the above scenario, i.e. given this specific triangle but with $q,r$ variable?
5. Can you perhaps even select $\delta$ such that it is independent of the triangle in question? Wikipedia might provide some suggestions (thanks to one question here, which also relates to another question here).
• @LiamMarsh: Item 5 suggests there might be a way to find a $\delta$ independent of the triangle, i.e. a constant of the hyperbolic plane itself. The links provided there will help you answer this question yourself. Up to item 5 I'd say not depending on $q,r$ should be enough, to get you started. But that's just intermediate steps. – MvG Dec 29 '17 at 12:30
• @LiamMarsh: You are on the right track. Although you might simplify things by noting that you are not asked to choose the bound $\delta$ tight, i.e. as low as possible, which is what $\ln(1+\sqrt2)$ would be. So perhaps the proof would be easier by observing that $p_b,p_c$ are both on the incircle and any points on the incircle are at most twice its radius apart? $d_{\mathbb D}(q,r)\le d_{\mathbb D}(p_b,p_c)$ feels indeed both true and crucial. How to show that depends a bit on your background, whether to focus on trigonometry or cross ratios or something else. – MvG Dec 29 '17 at 17:00
• @LiamMarsh: It isn't, in hyperbolic geometry. The closer $q$ and $r$ are to $a$, the smaller the triangle $\triangle aqr$ and the smaller the angle deficit. In fact you have $\angle arq \ge \angle ap_bp_c$ with equality only for $q=p_c$. But if you look at e.g. the hyperbolic law of sines, you find that an angle increase as the edge lengths decrease will only strengthen your inequality. – MvG Dec 30 '17 at 17:51