# Show that any side of a hyperbolic triangle in the hyperbolic plane lies in the closed $δ$-neighborhood of the union of the two other sides.

Prove that there exists a constant $δ ∈ [0,∞)$ such that the following holds. Let T be an arbitrary non-degenerate hyperbolic triangle in $\mathbb D$ with vertices $a, b, c ∈ \mathbb D$, and 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 $d_\mathbb D(c, p_a) = d_\mathbb D(c, p_b)$

Then for any points $q ∈ [a, p_c]$ and $r ∈ [a, p_b]$ with $d_\mathbb D(a, q) = d_\mathbb D(a, r)$ we have $d_\mathbb D(q, r) ≤ δ$

Conclude that any side of a hyperbolic triangle in the hyperbolic plane lies in the closed $δ$-neighborhood of the union of the two other sides.

Can I just say the following?

Let $δ = 2max[d_\mathbb D(a, p_b),d_\mathbb D(b, p_c),d_\mathbb D(c, p_a)]$.

As $d_\mathbb D(x, y)$ is defined to be the infimum of the length of all paths between $x$ and $y$, $d_\mathbb D(r, q) \le d_\mathbb D(a, r) + d_\mathbb D(a, q) \le d_\mathbb D(a, p_b) + d_\mathbb D(a, p_c) = 2d_\mathbb D(a, p_b)$.

You get similar results for the other 2 vertices and hence I choose $δ$ to be the maximum of $2d_\mathbb D(a, p_b), 2d_\mathbb D(b, p_c)$ and $2d_\mathbb D(c, p_a)$.

• I'm sorry but I don't understand. Could you point me in the right direction please? – user2973447 Jan 3 '17 at 23:13
• see en.wikipedia.org/wiki/%CE%94-hyperbolic_space , ( but your construction is a bit more limited ) – Willemien Jan 9 '17 at 18:11
• Sorry I'm finding that page a bit hard to understand. Why is my construction more limited? It seems to me that from that page any side of my hyperbolic triangle will lie in the δ-neighbourhood of the other 2 sides. Thanks for your help with this! – user2973447 Jan 17 '17 at 3:21
• In your construction you presume that the point nearest to $p_c$ on any side is the point $p_b$ on $ac$ this ignores the option that a point on $bc$ can be closer to $q$ than $p_b$ and this can be the case if $\angle bac$ is greater than right. maybe study en.wikipedia.org/wiki/Hyperbolic_triangle and en.wikipedia.org/wiki/Ideal_triangle it all follows that there are limits to the radius of the incircle. – Willemien Jan 17 '17 at 21:48

I am not sure at the moment if the below is true treat with care and be critical

OLD VERSION (no new one yet)

After some thinking I realised the answer is no. there is no such constant.

from $d(a, p_b) = d(a, p_c)$, $d(b, p_a) = d(b, p_c)$ and $d_(c, p_a) = d(c, p_b)$

you can calculate the distance $d(a,p_b) = d(a,p_c) = \frac{1}{2} \left( d(a,b)+ d(a,c)-d(b,c) \right)$ and has no maximum

Then Suppose $\angle a \ge$ right angle then $a$ is the closest point on $ap_c$ to $ap_b$ so $d(q,r) > d(a,r)$

Also suppose the point $r$ is the point $P_b$

The distance $d(a,r)$ has no maximum so also the distance $d(q,r)$ has no maximum

so there is no such constant.

I think you make the mistake that the shortest distance of q to any of the edges is to a point on $ab$ while even with the conditions you place on $q$ a point on $bc$ can be closer.

hope this helps