Here's a curious set of vertices for a tetrahedron: {{-22, -25, 4}, {-12, 15, -6}, {8, 5, -6}, {18, -15, 24}}

The Fermat point of a tetrahedron minimizes the total distances from the point to the vertices. There is a general method for finding the Fermat point of polygons using the Weiszfeld algorithm.

The six edge lengths of this tetrahedron are different. The Fermat point is at the origin.

Find a second tetrahedron with six different edge lengths, integer coordinates away from the origin, and a Fermat point at the origin. The second tetrahedron should not be a variation of the first tetrahedron.

For thousands of other differently-edged integer-vertex tetrahedra I've looked at, the Fermat point requires the algebraic roots of three sextic to octic polynomials. And then there's this one integer solution.

For a triangle, the projection of the vertices onto a circle centered on the Fermat point splits the circle into 3 equal arcs.

For a tetrahedron, a projection of the vertices onto a sphere centered on the Fermat point splits the sphere into 4 equal spherical triangles. (Is this known?)

special tetrahedron

Incenter - tiny sphere near Fermat
Centroid - green {-2, -5, 4}
Circumcenter - cyan {-(33/4), -(5/2), 71/4}
Twelve point - yellow {1/12, -(35/6), -(7/12)}
Symmedian - pink {42/89, 315/89, -(144/89)}
Fermat - red {0,0,0}
Monge - magenta {17/4, -(15/2), -(39/4)}

This diagram made with code from Tetrahedron Centers


Any tetrahedron works with vertices of the following form:





with $a,b,c,d$ all the same sign. The "radii" from the origin to each vertex form "tetrahedral angles" with each other making the origin the Fermat point. If we require integer solutions, we can of course require $a,b,c,d$ to be appropriate same-signed quantities and apply a suitable rotation.

Let us look at the given solution. Imagine that the first vertex is a rotation of $(a,a,a)$ as rendered above. Then $3a^2=22^2+25^2+4^2=1125=9×125$. Similarly $b^2=375=3×125, c^2=125, d^2=1125$. Note that $a,b,c,d$ are not necessarily rational multiples of each other, although (at least here) their squares are.

We can, of course, construct simpler solutions that do not require rotations, like $(1,1,1),(2,-2-2),(-3,3,-3),(-4,-4,4)$.

  • 1
    $\begingroup$ Such a construction only admits four degrees of freedom, instead of a general tetrahedron's full complement of six. So, this is an interesting sub-family, like orthocentric tetrahedra. As with the orthocentrics, these theoretically admit parameterization by face areas. Expressing $a$, $b$, $c$, $d$ in terms of those areas doesn't seem to be easy, though. $\endgroup$ – Blue Jul 28 '18 at 21:00

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