Find (for n=2,3,4,5,6,7,8,9,10) a configuration of n points on a regular tetrahedron ABCD such that the minimum distance between two points is as great as possible.

This year, the Dutch professionals couldn't beat an American: mister Jack Burns from Los Angeles won the main prize. Proficiat, Jack!
The Dutch participants from Tilburg and Maastricht did find the optimal configurations for n=2,3,4,5,7,8,9,10.
However, for n=6 they stuck to a non-optimal configuration, deducted from the case n=8 by omitting two points. Evidently, they didn't imagine in this case that two pirates should move a bit closer to each other from their positions in a vertex of the tetrahedron.
Nevertheless: thanks for your contributions.

The winning contribution of Jack Burns can be found here. It is a bit concise, but without any errors.

The prize for amateurs goes to Johan van der Meer from Monnickendam. Congratulations, Johan!
He works with an unrolled tetrahedron as well, using rotation symmetry. He uses both polar coordinates and a coordinate system that is adapted to the tetrahedron. For n=2,3,4,5,9,10, he finds the optimal configurations.
He erroneously deducts his configurations for n=6,7,8 from the one for n=10 by omitting points. If you look at the centers of the faces, you find slightly better configurations!
He made a little error with the theorem of Pythagoras. That's why he didn't always find the exact minimal distances with his optimal configurations, either.
But he did find six out of the nine optimal configurations!

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