A regular tetrahedron is centered at the origin with vertices $ (0,0,1$) and $(a,0,b)$. All of its vertices satisfy $x^2+y^2+z^2=1$
Find the remaining two vertices with respect to $a$ and $b$.

Let $c=\dfrac{\sqrt{2}}{3}$, $d=\dfrac{\sqrt{2}}{\sqrt{3}}$
With these notations, the solution is: 
$A_1(0,0,1)$ (given), $A_2(2c,0,-1/3)$, $A_3(-c,d,-1/3)$ and $A_4(-c,-d,-1/3)$
The way I have done it: 
1) As the center of mass is necessarily the origin, knowing that a regular tetrahedron has its center of mass at the $1/4$ of its height, points $A_2, A_3, A_4$ constituting the (horizontal) base of the tetrahedron must all have their $z$ coordinate equal to -1/3.
In particular $b=-1/3$.
2) Obtaining $a$: because of the sphere constraint $a^2+0^2+(1/3)^2=1$, we have $a=2c$.
3) Obtaining edge length (squared) : Knowing now the coordinates $A_1$ and $A_2$, the value of the square of the length of any edge is $(A_1A_2)^2=8/3$.
4) Let $A_3=(e,f,-1/3)$ and $A_4(g,h,-1/3)$. Then "sphere constraints" give 
$e^2+f^2+1/9=1 \ \ (1)$ and $g^2+h^2+1/9=1 \ \ (2)$. 
5) Edge lengths constraints (as seen before) i.e., $(A_1A_3)^2=8/3$ and $(A_1A_4)^2=8/3$ give equations $(e-1)^2+f^2+1/9=8/3  \ \ (3)$ and $(g-1)^2+h^2+1/9=8/3 \ \ (4)$.
6) Using equations (1), (2), (3), (4) the four unknowns $e,f,g,h$ are easily obtained.
Remark: if you know rotation matrices, after items 1) and 2) one can proceed to a computation that applies the following rotation matrix (axis $0z$, angle $2 \pi/3$) twice to vector $\vec{OA_2}=\begin{bmatrix}2\sqrt{2}/3\\0\\-1/3\end{bmatrix}$:
$$R=\begin{bmatrix}-1/2& -\sqrt{3}/2& 0\\\sqrt{3}/2& -1/2& 0\\0& 0& 1\end{bmatrix}$$
