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I am considering a unit sphere with center at $(0, 0, 0)$ and spherical triangles congruent to the one with the vertices in $(1, 0, 0)$, $(0, 1, 0)$ and $(0, 0, 1)$. If you allow for triangles to have overlapping sides, then it's obvious that you can put at most $8$ of them. What happens if I want them to be completely disjoint? What is the maximal number of triangles one can fit on a sphere?

Guided by intuition, I thought you could put $7$ of them. But, the best I could do with a bit of imagination is $4$.

If the question appears to be too trivial, I am interested in generalisations in higher dimensions as well. Any ideas or references are appreciated.

prosinac
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    I can think of 5. Take the triangle T with the vertices in (1,0,0), (0,1,0) and (0,0,1), three its neigbouring triangles, and a triangle opposite to T. Rotate the latter around its center. Now you have place to move three neighbors of T away from T and from each other. – colt_browning Sep 19 '17 at 22:04
  • that's still 4, but I get what you mean, I can do 5 now – prosinac Sep 21 '17 at 18:43

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