Hyperbolic Triangles and Uniform thinness

Question

My textbook states that all triangles in hyperbolic space are uniformly thin in the following way:

If $ABC$ is a triangle and $x$ is a point on one side, then there exists a point $y$ on one of the other two sides such that $d(x,y) \leq \ln(1 + \sqrt{2})$.

I am trying to get an intuitive grasp on this phenomenon, as well as a formal proof. Thinking about the Poincare disk model of the Hyperbolic Plane, the geodesic between points $A$ and $B$ will be the arc of the circle connecting them that is orthogonal to the boundary of the disk. It seems to me that the case to worry about the most is when the point $x$ is in the middle of one of the sides. I can kind of imagine how sliding a point $x$ along such a geodesic (or side of the triangle) will take it closer to the center of the disk and thus closer to the other sides as well. However, this approach does not hint towards a formal solution.

Answer 1

I don't think moving to the disk model immediately is a good way to think of this.

Instead, construct a normal to the side at $X$. This normal will intersect one of the two other sides at some point $P$. Let's assume that $X$ is on $BC$ and the normal intersects $AB$. (If the normal happens to go through $A$ exactly, you get a choice of which side to work with). It is reasonably to expect that a close approach of one of the other sides will be somewhere on $AB$, then. So it will be enough to prove

Suppose $XPB$ is a right triangle, with $X$ being the right angle. Then some point on $PB$ will be at most $\log(1+\sqrt 2)$ from $X$.

To see this, consider the line $m$ that is parallel to $XP$ in one direction and $XB$ in the other. The point $M$ on $m$ that is closest to $X$ can be found by drawing a perpendicular to $m$ from $X$.

Now, no matter where on the lines $P$ and $B$ are, the line segment $PB$ must intersect $XM$ somewhere. (If it goes farther from $X$ it would have to intersect $m$ two times, which is not allowed). So the intersection of $XM$ with $PB$ is closer to $X$ than $M$ is.

And thus, all you have to prove is that the altitude of a 90-0-0 triangle is $\log(1+\sqrt 2)$. (Which I suppose is true because that's what makes the textbook's claim work).