TITLE: Packing regular tetrahedra of edge length 1 with a vertex at the origin in a unit sphere
QUESTION [11 upvotes]: I'm currently venturing through Paul Sally's Fundamentals of Mathematical Analysis. This is an unusual textbook in terms of the difficulty of exercises. I've already been stunned by the very first one: 
How many regular tetrahedra of edge length 1 can be packed inside a unit sphere with each one has a vertex located at the origin? 
My instinct told me this is highly nontrivial, as I've struggled with it for several weeks. Could somebody give a hint on how to tackle it? Thanks in advance.
Edit: As jmerry and Erik Parkinson shows, the solution has been squeezed between $20$ and $22$. Can further progress be made?

REPLY [3 votes]: This is not an answer, but an extended comment.
The edge length of an unit icosahedron (all vertices at unit distance from origin) is $$a = \frac{1}{\sin\frac{2 \pi}{5}} \approx 1.0514622$$
according to the Wikipedia Regular icosahedron article. If we compare the net of such an icosahedron to twenty equilateral triangles with edge length 1 (corresponding to the faces of the tetrahedra opposite to the vertex at origin),

it is rather obvious that there is not enough room to squeeze in another tetrahedron in there. (This corresponds to actually making 20 unit tetrahedra, and trying to squeeze in an additional tetrahedron while keeping them all within an unit sphere. Remember, the above are nets, not a solid angle diagram.)
My math-fu is not advanced enough to express that as a convincing mathematical argument, though.

It might be interesting to note that R. H. Hardin, N. J. A. Sloane and W. D. Smith list the putative covering radii of points on sphere as (excerpt)
$$\begin{array}{c|c}
\text{points} & \text{degrees} \\
\hline
19 & 30.3686748 \\
20 & 29.6230958 \\
21 & 28.8244768 \\
22 & 27.8100588 \\
23 & 27.4818687 \\
\end{array}$$
This means that if we take right circular cones with apex angle say $2 \cdot 29.6230958 = 59.2461916$ degrees, up to twenty of them can be connected by their apexes; if the angle is say $2 \cdot 30.3686748 = 60.7373496$ degrees, only 19. 
Unfortunately, the incircle and circumcircle of the outer face of the tetrahedron differ too much to allow a stricter bound (than 20 to 22 tetrahedra).