Compare the heights of two midpoints in a hyperbolic triangle

Question

Let $\Delta ABC$ be a triangle in Hyperbolic Geometry. Suppose that the midpoints of $AB$ and $AC$ are $E$ and $F$ respectively. Then can we compare the their distances to the base $BC$? In other words, can we compare the "heights" of these two midpoints? Are they equal?

I guess they are related to length of $AB$ and $AC$, but I am not very clear about the precise relation.

Answer 1

With a slight change in notation, consider this figure:

where $b := \frac12|\overline{AC}|$ and $c := \frac12|\overline{AB}|$.

All we need is the Law of Sines and the Double-Angle Formula for Sine.

$$\begin{align} \triangle ABC\;:&\quad \frac{\sin B\;}{\sin C\;} = \frac{\sinh 2b}{\sinh 2c} = \frac{2\sinh b \cosh b}{2\sinh c\cosh c} \quad\to\quad \frac{\sin B \sinh c}{\sin C \sinh b} = \frac{\cosh b}{\cosh c} \\[4pt] \triangle BPP^\prime:&\quad \frac{\sin B\;}{\sin P^\prime} = \frac{\sinh p}{\sinh c} \quad\to\quad \sin B \sinh c = \sinh p \\[4pt] \triangle CQQ^\prime:&\quad \frac{\sin C\;}{\sin Q^\prime} = \frac{\sinh q}{\sinh b} \quad\to\quad \sin C \sinh b = \sinh q \end{align}$$ Thus,

$$\frac{\sinh p}{\sinh q} = \frac{\cosh b}{\cosh c} $$

As one might expect, the "heights" $\overline{PP^\prime}$ and $\overline{QQ^\prime}$ are congruent if and only if $\triangle ABC$ is isosceles with vertex $A$. $\square$

Answer 2

I'll call the angles $\alpha$ at $A$, $\beta$ at $B$, and $\gamma$ at $C$ . The law of cosines gives us the angles in terms of the edge lengths:

$$\cosh AC = \cosh AB\cosh BC - \sinh AB\sinh BC\cos\beta$$ $$\cos\beta = \frac{\cosh AB\cosh BC - \cosh AC}{\sinh AB\sinh BC}$$

and similarly for $\alpha$ and $\gamma$ .

Construct a line $\overline{EG}$ perpendicular to the base $\overline{BC}$, with $G$ on $\overline{BC}$. Then apply the law of sines to the right triangle $\Delta BEG$ :

$$\frac{\sinh EG}{\sin\beta} = \frac{\sinh BE}{\sin\frac\pi 2} = \sinh BE$$

Squaring and applying trig identities,

$$\sinh^2 EG = \sinh^2 BE\sin^2\beta$$ $$= \frac{\cosh(2BE) - 1}{2}(1 - \cos^2\beta)$$

Point $E$ is the midpoint of $\overline{AB}$, so $2BE = AB$ . And we can use that other equation for $\cos\beta$ :

$$\sinh^2 EG = \frac{\cosh AB-1}{2}\left(1-\left(\frac{\cosh AB\cosh BC-\cosh AC}{\sinh AB\sinh BC}\right)^2\right)$$

And that's it! You can simplify or expand this a little, but this gives the height of $E$ in terms of the original triangle's edge lengths. And by symmetry, replacing $\{B,E,G,\beta\}$ with $\{C,F,H,\gamma\}$,

$$\sinh^2 FH = \frac{\cosh AC-1}{2}\left(1-\left(\frac{\cosh AC\cosh BC-\cosh AB}{\sinh AC\sinh BC}\right)^2\right)$$

A sample calculation shows that $EG$ and $FH$ are not equal in general. I used $AB=3$, $AC=4$, $BC=5$; then $\sinh^2 EG \approx 0.28$ , but $\sinh^2 FH \approx 0.11$ . But they are approximately equal with small distances: $AB=0.03$, $AC=0.04$, $BC=0.05$; then $\sinh^2 EG \approx 0.00014396$, and $\sinh^2 FH \approx 0.00014394$ .

Answer 3

A proof without any real computations (just some intuition about hyperbolic spaces) that they cannot be equal:

An important property of hyperbolic spaces is that all triangles are $\delta$-thin, i.e., there exists a $\delta$ such that any point on the edge $AC$ is in distance at most $\delta$ from one of the edges $AB$ or $BC$. This means that hyperbolic spaces are very similar to trees (see e.g. Wikipedia for more background and details).

Now, consider a triangle $ABC$ where $AB=AC=d$ and the angle at $B$ is a right angle. The midpoint of $AB$ ($E$) will be in distance $d/2$ from the line $BC$, since the angle $ABC$ is right. Now, the midpoint of $AC$ ($F$) is in distance not more than $\delta$ from either $AB$ or $BC$; however, because of symmetry, its distances to $AB$ and $BC$ are equal, thus, the distance of $F$ from $BC$ is not more than $\delta$. Since $d$ can be arbritrarily large, we have $d(F,BC) < \delta < d < d(E,BC)$.

(If you are more familiar with the Poincaré disk model than $\delta$-thin triangles, this can be also seen quite easily by putting $B$ in the center of the disk, and also a very big right isosceles triangle.)