How to find a tetrahedron that covers four points?

Question

I’m looking for an explicit formula for the vertices of a regular tetrahedron that covers four given points. In particular:

Given four distinct real numbers $a_1$, $a_2$, $a_3$, $a_4$, is there a simple formula for four complex numbers $z_1$, $z_2$, $z_3$, $z_4$ such that the four points $(a_i,z_i)$, $i=1,\ldots,4$, in ${\Bbb R}\times{\Bbb C}$ form the vertices of a regular tetrahedron? That is, $|a_i-a_j|^2+|z_i-z_j|^2$ is independent of choice of distinct $i$, $j$.

Same question but with $a_i$ complex.

As an example of the type of thing I’m looking for, for three real numbers numbers $a$, $b$, $c$, $(a,(b-c)/\sqrt 3), (b,(c-a)/\sqrt 3), (c,(a-b)/\sqrt 3)$ are the vertices of an equilateral triangle.

I've tried things like $(a,(b+c\omega+d\overline\omega)/2)$ (where $\omega$ is a third root of unity) which looks to be on the right track but... I just hope someone knows the answer.

Thanks in advance.

Answer 1

I am not sure about the "simple" aspect, but if you replace $\mathbb{C}$ by $\mathbb{R}^2,$ what you are looking for is quadruple of vectors $z_0, \dotsc, z_3$ such that $$\|z_i - z_j\|^2 = C - (a_i - a_j)^2 = d_{ij}.$$ You can assume that $z_0 = 0,$ and then by using the parallelogram law, you have $G(z_1, z_2, z_3) =E,$ where $E$ is a matrix whose coefficients are linear functions of the $d_{ij},$ and $G$ is the Gram matrix. Now, for the solution to exist, $E$ has to be positive semidefinite. Once it is, you diagonalize it as $E = P D P^t,$ at which point $Z = P \sqrt{D} P^t.$ Whether or not there is a choice of $C$ such that $E$ is positive semidefinite is a separate question which can be addressed by computing the characteristic polynomial of $E$ and checking that it has the right kind of roots. This will give you potentially unpleasant equations satisfied by $C,$ but I haven't done the computation, and you should...

Answer 2

A regular tetrahedron is inscribed in a cube. Finding a regular tetrahedron whose vertices project to $a_1,a_2,a_3,a_4$ is equivalent to finding a cube so that one vertex projects to $a_0 = \frac{1}{2}(a_1+a_2+a_3-a_4)$ and three adjacent vertices project to $a_1, a_2, a_3$.

Translate the values so that $a_0 = 0$ (equivalently, translate so that $a_4 = a_1+a_2+a_3$). We want $3$ orthogonal vectors of equal length whose first coordinates are $a_1, a_2, a_3$. Rescale so that $a_1^2+a_2^2+a_3^2=1$. Equivalently (transposing), we want an orthogonal matrix whose first column is $\vec v =\begin{smallmatrix}a_1 \\a_2 \\a_3\end{smallmatrix}$. For example, we can reflect $\vec e_1=\begin{smallmatrix} 1 \\ 0 \\ 0\end{smallmatrix}$ to $\vec v$ (assuming $\vec v \ne \vec e_1$). Let $\vec w = \|\vec v - \vec e_1 \|^{-1} (\vec v - \vec e_1) = \begin{smallmatrix} w_1 \\ w_2 \\ w_3 \end{smallmatrix}.$ The reflection $\vec u \mapsto \vec u - 2(\vec u \cdot \vec w) \vec w$ corresponds to the orthogonal matrix

$$\begin{pmatrix} a_1 & -2w_1w_2 & -2 w_1w_3 \\ a_2 & 1-2w_2^2 & -2 w_2w_3 \\ a_3 & -2w_2w_3& 1-2w_3^2\end{pmatrix}.$$

So, the points $(a1,-2w_1w_2,-2w_1w_3)$,$(a_2,1-2w_2^2,-2w_2w_3)$,$(a_3,-2w_2w_3,1-2w_3^2)$, and $(a_1+a_2+a_3, 1-2w_2(w_1+w_2+w_3),1-2w_3(w_1+w_2+w_3))$ are the vertices of a regular tetrahedron.