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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.

Martin Welsh
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    You certainly can get at least $2$ points - one a relfection of the other in the plane of the three given points. Whether there might be more depends on what you mean by "given all side lengths." If you don't know which lengths go where, it is possible to get multiple points. For example, if the three given points are in an equilateral triangle, then you get $6$ points. – Thomas Andrews Oct 10 '17 at 02:36
  • I probably should have been more clear in my question: the lengths of all sides are known (ie, for tetrahedron ABCD; AB, AC, AD, BC, BD, and CD are known). Also, I meant not just if finding the position of the fourth vertex can be done, but how to find the position of the fourth vertex – Martin Welsh Oct 10 '17 at 02:56
  • Edit your question, then. Answerers should not be required to read comments to understand the question. Help people help you. – Thomas Andrews Oct 10 '17 at 02:58
  • thanks for the help – Martin Welsh Oct 10 '17 at 03:17
  • Very cool question! With AB, BC and AC, I can get a base of the tetrahedron. With BD, I get a half-sphere around B as the locus of D. With CD, I get another half-sphere around C. They intersect, and then with AD one hopefully gets a unique solution. I'm excited to translate this all into algebra. Thanks for the problem! Reminds me of Polya's "Pattern of Two Loci." – MBP Oct 10 '17 at 03:26

2 Answers2

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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.

MBP
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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 :)

mgold
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