Related to angles in tetrahedron

Question

Let $OABC$ be a tetrahedron such that $|OA|=|OB|=|OC|$. Denote by $D$ and $E$ the midpoints of segments $AB$ and $AC$ respectively. If $\alpha=\angle(DOE)$ and $\beta=\angle(BOC)$ what is the ratio $\beta/\alpha$?

It is obvious that $|BC|=2|DE|$ since triangles $\Delta(ABC)$ and $\Delta(ADE)$ are similar and $D, E$ are midpoints by assumption. I would expect that $\beta/\alpha\geqslant 2$ but I haven't been able to confirm this. I tried to use the cosine law for the segments $|BC|$ and $|DE|$ but without success.

Answer 1

The ratio $\beta/\alpha$ does not take a single value. This answer does not attempt to find a formula for the ratio.

I imagine this becomes finding the distance of two mid-points of a spherical triangle.

Construct a sphere centred at $O$ with radius $r=OA$.

Extend $OD$ and $OE$ to meet the sphere at $D'$ and $E'$ respectively; $D'$ and $E'$ will be the mid-points of arcs $\overset{\frown}{AB}$ and $\overset{\frown}{AC}$ respectively.

Then the required angles $\angle BOC$ and $\angle D'OE'$ are proportional to the lengths of arcs $\overset{\frown}{BC}$ and $\overset{\frown}{D'E'}$ respectively:

$$\frac\beta\alpha = \frac{\angle BOC}{\angle D'OE'} = \frac{\overset{\frown}{BC}}{\overset{\frown}{D'E'}}$$

Now consider four special / extreme cases:

Case 1: When $A$, $B$ and $C$ are near relative to the size of the sphere, Euclidean geometry becomes applicable, and arcs and respective chords are approximately equal in length. By mid-point theorem,

$$\overset{\frown}{BC} \approx 2\overset{\frown}{D'E'},\quad \frac\beta\alpha \approx 2$$

Case 2: When $A$ is the north pole of the sphere, and $B$ and $C$ on the equator. $D'$ and $E'$ are on the $45^\circ N$ latitude. Comparing the arc lengths of $\overset{\frown}{BC}$ and $\overset{\frown}{D'E'}$,

$$\overset{\frown}{BC} >\overset{\frown}{D'E'},\quad \frac\beta\alpha > 1$$

Case 3: When $A$ is the north pole, and $B$ and $C$ are on the $30^\circ S$ latitude. $D'$ and $E'$ are on the $30^\circ N$ latitude, with the same longitude as $B$ and $C$ respectively. Then $\overset{\frown}{D'E'}$ and $\overset{\frown}{BC}$ are symmetric, reflected by the equatorial plane,

$$\overset{\frown}{BC} = \overset{\frown}{D'E'},\quad \frac\beta\alpha = 1$$

Case 4: When $A$ is the north pole, and $B$ and $C$ are close to the south pole. The spherical triangle of $ABC$ is almost like a spherical lune. $D'$ and $E'$ are just north of the equator.

Arc $\overset{\frown}{BC}$ can be arbitrarily short, while $\overset{\frown}{D'E'}$ depends on the longitudes of $B$ and $C$ and tends to $r$ times the difference in longitudes.

$$\overset{\frown}{BC} \ll \overset{\frown}{D'E'},\quad \frac\beta\alpha \approx 0$$

Answer 2

This answer attempts to find a formula for the ratio.

I imagine this becomes finding the distance of two mid-points of a spherical triangle.

Construct a sphere centred at $O$ with radius $r=OA$.

Extend $OD$ and $OE$ to meet the sphere at $D'$ and $E'$ respectively; $D'$ and $E'$ will be the mid-points of arcs $\overset{\frown}{AB}$ and $\overset{\frown}{AC}$ respectively.

Following this convention of spherical trigonometry, consider the spherical triangle $ABC$, denote

$$\begin{align*} a &= \angle BOC = \beta\\ b &= \angle AOC\\ c &= \angle AOB \end{align*}$$

According to the spherical laws of cosine, if $A$ is the spherical angle made by arcs $\overset{\frown}{AB}$ and $\overset{\frown}{AC}$, consider spherical triangle $ABC$,

$$\begin{align*} \cos A &= \frac{\cos a -\cos b \cos c}{\sin b \sin c}\\ &= \frac{\cos \beta -\cos b \cos c}{\sqrt{(1+\cos b)(1-\cos b)(1+\cos c)(1-\cos c)}}\tag 1\\ \end{align*}$$

And consider spherical triangle $AD'E'$,

$$\begin{align*} \cos A &= \frac{\cos \angle D'OE' -\cos \frac b2 \cos \frac c2}{\sin\frac b2 \sin\frac c2}\\ &= \frac{\cos \alpha -\sqrt{\frac{1+\cos b}2} \sqrt{\frac{1+\cos c}2}}{\sqrt{\frac{1-\cos b}2}\sqrt{\frac{1-\cos c}2}}\\ &= \frac{2\cos \alpha - \sqrt{(1+\cos b)(1+\cos c)}}{\sqrt{(1-\cos b)(1-\cos c)}}\tag 2 \end{align*}$$

Equating $\cos A$ in $(1)$ and $(2)$,

$$\begin{align*} \frac{\cos \beta -\cos b \cos c}{\sqrt{(1+\cos b)(1+\cos c)}} &= 2\cos \alpha - \sqrt{(1+\cos b)(1+\cos c)}\\ \cos \beta -\cos b \cos c &= 2\cos \alpha\sqrt{(1+\cos b)(1+\cos c)} - (1+\cos b)(1+\cos c)\\ \cos \beta &= 2\cos \alpha\sqrt{(1+\cos b)(1+\cos c)} - 1 - \cos b - \cos c\\ \cos \alpha &= \frac{1 + \cos b + \cos c + \cos \beta}{2\sqrt{(1+\cos b)(1+\cos c)}}\\ &= \frac{1 + \cos b + \cos c + \cos \beta}{4\cos\frac b2 \cos \frac c2}\\ \frac \beta\alpha &= \frac{\beta}{\arccos \cfrac{1 + \cos b + \cos c + \cos \beta}{4\cos\frac b2 \cos \frac c2}} \end{align*}$$