geometry
What kind of regular tetrahedra can we inscribe into a given torus?
For example, is it possible to inscribe a unit edge length regular tetrahedron to a spindle torus with major radius $R=\frac{\sqrt 2}2$ and minor radius $r=\frac{\sqrt 3}2$?
It seems to me that such an inscription should be doable for any $R$ and $r$ in a reasonable range, but I'm having a very hard time imagining it; I hope someone with some knowledge/vision in spatial geometry can answer this from the top of their head.
If the side length is 1, then the height of a face is $\sqrt3\over2$ by Pythagoras. Let $P$ be the projection of $B$ onto $ACD$. By symmetry it is the center of $ACD$, hence at a distance of $\frac23\cdot {\sqrt3\over2}={\sqrt3\over3}$ from $A$. Then ${\sqrt3\over3}$ is also the cosine of $\angle PAB$. Thus $\angle PAB = \arccos({\sqrt3\over3}) \approx 0.9553166181245 \approx 54.7356103172^o$.
As already pointed out in a comment by Moishe Cohen and especially
the comment by Alex Ravsky, the answer is negative. Let attention
be restricted to the regular tetrahedron: dividing the length of each edge by two would result in $2^3 = $ eight smaller instances
of that regular tetrahedron. Suppose that we try to assemble the original shape with these smaller tetrahedrons, placing four of
the $1/8$ ones at the corners of the original to begin with. Then there will be a void in the middle which has the shape of a regular
octahedron. It is impossible to fill that void with the remaining four smaller tetrahedrons. I have no rigorous proof for this,
but that doesn't serve as a disclaimer. Just use your imagination and you shall see:
The simplified 2-D case is shown in the picture on the left. The picture on the right is the top view of a 3-D wire frame model of the
subdivided regular tetrahedron. The four smaller ones at the corners of the original shape ABCT are: AcbP , BacQ , CbaR at the bottom and PQRT at the first floor.
Best Answer
One solution has the vertices: $$\begin{array}{ccc} (-0.42, & \ \ \, 0.28, & 0.84 )\\ (-0.03, & -0.16, & 0.04 )\\ (-0.19, & -0.69, & 0.87 )\\ (\ \ \, 0.54, & \ \ \, 0.00, & 0.85 )\\ \end{array}$$ which lead to this view looking up at the region with $z>0$:
This comes from Mathematica using NMinimize, which surprisingly worked better than NSolve or FindInstance. The code is below, and you could get other solutions by rotating this or by adding a term like $(a_2-\frac32)^2$ to the sum of squares.