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?


  • $\begingroup$ 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. $\endgroup$ – Paul Sinclair Dec 28 '17 at 23:47
  • $\begingroup$ 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. $\endgroup$ – 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:


  1. Can you describe the relevant parts of this configuration in words?
  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?
  3. Can you prove it?
  4. Can you adjust the proof so it still holds in hyperbolic geometry?
  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).
  • $\begingroup$ @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. $\endgroup$ – MvG Dec 29 '17 at 12:30
  • $\begingroup$ @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. $\endgroup$ – MvG Dec 29 '17 at 17:00
  • $\begingroup$ @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. $\endgroup$ – MvG Dec 30 '17 at 17:51
  • $\begingroup$ @LiamMarsh: The fact that you deleted your comments makes it hard for future visitors to follow your insights. I encourage you to write your own answer, expressing your understanding in your own words. My question was meant to guide you, not answer you straight away. But future visitors likely won't start a conversation to clarify unclear points. Thus a straight answer may be more useful in the long run. Out of curiosity, what made you delete your comments and remove the accept mark? $\endgroup$ – MvG Jan 2 '18 at 19:56

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