A regular tetrahedron is centered at the origin with vertices $ (0,0,1$) and $(a,0,b)$.

Question

All of its vertices satisfy $x^2+y^2+z^2=1$ Find the remaining two vertices with respect to $a$ and $b$.

Answer 1

To solve everything from scratch will result in a too long answer, so we assume these important properties to be known:

Image 1: Tetraeder in cube construction. (Source)

1. The tetraeder in the task has the internal arm length $1$, meaning the distance from origin to top vertex. Then the edge length between two vertices is $L=2\sqrt{2/3}$.
   (This corresponds to a cube length $s=2/\sqrt{3}$ in the above image)
2. The internal angle between two arms is $\theta\approx 109^0$ with $\arccos \theta = -1/3$.
   (This is only used to check the following results)

For the second vertex $Q=(a,0,b)$ we have the constraints that it lies on the unit sphere around the origin and on the sphere around $P=(0,0,1)$ of radius $L$. \begin{align} 1 &= x^2 + y^2 + z^2 \\ L^2 &= x^2 + y^2 + (z-1)^2 \end{align} The intersection is the circle at height $$ L^2 = 2-2 z \iff z = 1 - L^2/2 = 1-4/3 = -1/3 $$ with the equation $$ x^2+y^2=1-1/9=(2\sqrt{2}/3)^2 \quad (*) $$ which has radius $r = 2 \sqrt{2}/3$.

(Large version)

Image 2: Unit sphere constraint (red), $L$ sphere constraint (green). plane constraints $z=-1/3$ and $y = 0$.

Cut with the plane $y=0$ this leaves two choices: $$ Q=\left( \pm2\sqrt{2}/3,0,-1/3 \right) $$ Note that we can pick only one of these solutions and not both as feasible vertices, as the angle between them is not $\theta$. We continue with the positive choice $Q_1$.

The constraints for $P$ and $A$ again lead to two choices for the third vertex $R$:

(Large version left image) (Large version right image)

Images 3 and 4: Unit sphere (purple) and $L$ sphere constraints for $P$ (red) and $Q$ (green). Intersection circles (yellow).

The intersection of unit sphere and $L$ sphere of $Q$ $$ 1 = x^2 + y^2 +z^2 \\ L^2 = (x-2\sqrt{2}/3)^2 + y^2 + (z+1/3)^2 $$ gives $$ L^2 = 8/3 = 1 -(4\sqrt{2}/3)x + 8/9 + (2/3)z + 1/9 \iff \\ 1 = - 2\sqrt{2}x + z $$ as equation of the plane where the intersection circle lies in.

Intersecting this plane with circle $(*)$ we get the equations $$ 1 = -2\sqrt{2} x -1/3 \iff x = -\sqrt{2}/3 $$ and $$ 8/9 = 2/9 + y^2 \iff y = \pm \sqrt{6}/3 $$ which gives two feasible vertices $$ R = (-\sqrt{2}/3, \pm \sqrt{6}/3, -1/3) $$ It turns out that these are the remaing two vertices we are looking for.

(Large version)

Image 5: The resulting tetraeder $(P,Q_1, R_1, R_2)$. Note the two measured tetraeder angles $\theta$. Note the alternative tetraeder vertex $Q_2$.

Summary:

We found two tetraeder that fit the task. $$ \begin{array}{l} P = \left( 0, 0, 1 \right) \\ Q_1 = \left( 2\sqrt{2}/3,0,-1/3 \right) \\ R_! = \left( -\sqrt{2}/3, \sqrt{6}/3, -1/3 \right) \\ R_2 = \left( -\sqrt{2}/3, -\sqrt{6}/3, -1/3 \right) \end{array} \quad\quad \begin{array}{l} P = \left( 0, 0, 1 \right) \\ Q_2 = \left( -2\sqrt{2}/3,0,-1/3 \right) \\ R_3 = \left( \sqrt{2}/3, \sqrt{6}/3, -1/3 \right) \\ R_4 = \left( \sqrt{2}/3, -\sqrt{6}/3, -1/3 \right) \end{array} $$

Answer 2

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 $\vec{OA_2}=\begin{bmatrix}2\sqrt{2}/3\\0\\-1/3\end{bmatrix}$:

$$R=\begin{bmatrix}-1/2& -\sqrt{3}/2& 0\\\sqrt{3}/2& -1/2& 0\\0& 0& 1\end{bmatrix}$$