I am trying to solve an exercise where I have to find a tetrahedron in a unit sphere. My first thought and i think it is not a bad idea is to let one point be at $(0,1,0)$ (assume this is $A$) and then find a plane

$y = -d$

so that

$|\overline{AB}|$ = $|\overline{BC}|$

using a new point $O = (0, d, 0)$ which will be the centroid of the triangle $BCD$. Now i know that the points $B, C, D$ will be on the sphere so that the equation

$x^2 + z^2 = 1 - d^2$

holds true.
Furthermore, I know that

$2 \* sin(60) \* \overline{OB} = \overline{BC}$.

I feel like i have collected all the parts of the puzzle, but i still cannot solve it. Any suggestions?

EDIT: Since somone mentioned it in the comments, it should be a regular tetrahedron