Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|cite|improve this question
Why do you want your points to be in $\mathbb{R} \times \mathbb{C}$, and not in $\mathbb{R}^3$? – Stefan Kohl Mar 5 '14 at 18:12

2 Answers 2

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...

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.