Dividing spherical triangles on sphere into 4 self similar smaller spherical triangles?

Starting from an intersection of the vertices of a tetrahedron with a sphere;

Is it possible to recursively divide the 4 spherical triangles into 4*4 = 16 smaller triangles according to the pattern below?:

Is self-similarity preserved when dividing the spherical triangle into 4 equal spherical triangles? In other words, can you keep dividing the spherical triangles ad infinitum?

[EDIT:] Due to spherical Excess the angles of the smaller spherical triangle will be different than the larger spherical triangle. If a spherical triangle is defined by angles a,b,c, alpha, beta, gamma is it possible to find equations to determine these angles using the number of divisions? Inother words the number of times a large spherical triangle has been partitioned in 4 spherical triangles?

• It should be possible, but the smaller triangles are not similar to the larger one (if this term is even defined for triangles on the sphere), because their sum of internal angles will not be the same (this sum is directly linked to the area of the triangles). For the lare triangle, the sum is $270^\circ$, while for smaller triangles it tends to $180^\circ$. Apr 3, 2018 at 11:41
• Created a simplified version of this question math.stackexchange.com/questions/2720234/… Apr 3, 2018 at 12:08
• Correcting a mistake of my former comment, the sum of the internal angles in the large triangles is $360^\circ$. Apr 3, 2018 at 12:28

Let $u$ and $v$ be two points on the unit sphere. The midpoint of the arc between them is found from the midpoint in space $u+v\over2$, then normalizing it - stretching it to have unit length.
The points of a regular tetrahedron might be $$A(c,s,0), B(c,-s,0), C(-c,0,s), D(-c,0,-s)\\c=\sqrt{1\over3},s=\sqrt{2\over3}$$ All points are a distance $\sqrt{8/3}$ from each other.
The midpoints between the first three points of the tetrahedron are:

Midpoint of $AB$ is $(1,0,0)$;
Midpoint of $BC$ is $(0,-\sqrt{1\over2},\sqrt{1\over2})$;
Midpoint of of AC is $(0,\sqrt{1\over2},\sqrt{1\over2})$

These three points are perpendicular to each other, so the middle triangle when you quarter $\Delta ABC$ has three right angles. The other three have angles $120^\circ,45^\circ,45^\circ$.

• My impression is that OP is talking about a triangle drawn on the surface of the sphere, using segments of great circles, not the triangular face of the tetrahedron. Apr 3, 2018 at 16:33

It’s really not clear what you’re asking. Subdivide each of the four original triangles (call one of them $\triangle ABC$) into four by bisecting the sides, say the midpoint of great-circle segment $\overline{AB}$ is $c$, and similarly decompose to get bisected segments $\overline{BaC}$ and $\overline{CbA}$. Drawing the segments $\overline{ab}$, $\overline{bc}$, and $\overline{ca}$, you get four new little triangles $\triangle{abc}$, $\triangle{Abc}$, $\triangle{aBc}$, and $\triangle{abC}$.

Are you asking whether the four new ones are mutually congruent equilateral, equiangular triangles? But of course, by the SSS criterion.

Are you asking whether the process may be repeated without let or hindrance? Again, certainly.

In spite of what I said above, the middle triangle of the four new ones is not congruent to the others. It alone is equilateral, the triangles at the three corners are not. Thanks to Achille Hui for pointing this out.

I’ll try thinking this through.

• Oh damn. The middle triangle is not congruent to the others, of course! Thank you so much, @achillehui. I’ll delete my Addendum. Thanks once more. Apr 4, 2018 at 4:35