Irregular tetrahedron problem. If you are given all side lengths of an irregular tetrahedron (ie, for ABCD we know the lenght of AB and AC and AD and so on), and given the positions of three of it's vertexes - how do you find the position(s) of the fourth vertex.
 A: Consider first three of the segments -- $AB, BC, AC$ -- and their lengths:
$$AB = s,
BC = r,
AC = t,$$
Place them on a plane as follows:
$$A: (0,0, 0)$$
$$B: (s,0, 0)$$
Then, C's location can be found by solving a system of equations representing the locus of $AC$ and of $BC$:
$$x^2 + y^2 = t^2$$ 
$$(x - s)^2 + y^2 = r^2$$
For the sake of ease, let's say that point $C$'s coordinates turn out to be $(e,f, 0)$.
Likewise, we can set up a super-laborious system of equations for the remaining segments, $AD, BD, CD$ and their lengths:
$$AD = m, BD = n, CD = p$$
Those equations would all represent spheres around $A, B, C$, the radii the various lengths of $AD, BD, CD$ respectively. The intersection of any two spheres is a circle, the intersection of all three would be a single point $D$:
$$x^2 + y^2 + z^2 =m^2$$
$$(x - s)^2 + y^2 + z^2 = n^2$$
$$(x - e)^2 + (y-f)^2 + z^2 = p^2$$
3 equations, 3 unknowns, we'll find that intersection point representing point $D$ of the tetrahedron. And if your tetrahedron lives elsewhere in space, you can slide this one around until its coordinates are where you want them to be.
A: Assuming that D is the unknown point, draw a triangle with points A and B separated by a leg of length AB, and the other legs of length AD and BD. Drop a vertical from the vertex of the AD and BD lengths. Use the law of cosines to determine all the angles, and from there the length of the vertical, and where it intersects the AB length edge. This will give you a circle on which D must lie. Its radius is the length of the vertical, and to find its center, figure out the proportions of how the vertical cuts AB and linearly interpolate between the A and B's cartesian values.
Repeat this construction with another choice of two known points and you will be left with the intersection of two circles in 3D, a question that has already been answered on this site.
First answer on math.stack; please be gentle :)
