Given 6 distinct points in $3$-$D$ space, can the distances between $3$ of the points be determined if all other distances between points are known?

Question

In the figure below of points in $3$-$D$ space, suppose the lengths of the blue segments are all known. Is it possible to determine the lengths of the red segments? Each point is connected to every other point by a line segment. The triangle $XYZ$ is red (lengths unkown), but all other lengths are known (blue).

$6$ points in $3$-$D$ space. The points are labeled $A$, $B$, $C$, $X$, $Y$, and $Z$. Each point is connected to each other point by a line segment. The segments of the triangle $XYZ$ are colored red. The other twelve line segments are colored blue." />

My geometry is rusty (especially in $3$-$D$), so I am not sure where to begin this problem. I would think that since $12$ of the segments are known and only $3$ are unknown, it may be possible to set up a system of equations to solve for the unknown lengths. But I am stumped on how to even create the equations. I found some equations about tetrahedra, but there is not an obvious (to me) way to combine them to create a solvable system.

The two most similar questions I found by searching are this one and this one. The first has a similar premise, but in that problem, the lengths are only known between two desired points and many other arbitrary points. The distances between the arbitrary points in that question are unknown, but they are known in my question. The second has more information about the points in a coordinate system, but in my problem no coordinates are known, just the distances between points.

If this problem can not be solved in the general case, could we add some assumptions to make it solvable? E.g. that triangles $ABC$ and $XYZ$ do not intersect, no three points are co-linear, triangle $ABC$ is/is not co-planar/parallel to triangle $XYZ$, etc.

Answer 1

This is not possible.

Counterexample: Suppose the distances between A, B and C are such that we can somehow place them on the unit circle in the xy-plane. Further suppose $|AX|=|BX|=|CX|=\sqrt{d_X^2+1}$ and that similar conditions for Y and Z hold. Then X can either be at $(0,0,d_X)$ or at $(0,0,-d_X)$ and similarly for Y and Z. Depending on the choice of the sign if the z-coordinate for each of the points, the distances $|XY|, |YZ|, |ZX|$ will be different.

The general problem at play is the following: First of all, it is not an issue that you don't know any coordinates of A, B and C. As you are only interested in the distances between X, Y and Z, you can WLOG place A at the origin and B, C in the xy-plane. The triangle ABC is uniquely determined by its sidelengths, any possible rotations around the z-axis don't matter for the desired lengths. Everything you know about X, for example, is that it lies on a sphere with radius $|XA|$ centered at A, on a sphere with radius $|XB|$ centered at B and on a sphere with radius $|XC|$ centered at C. However, in general, three spheres intersect in two points, either of which could be X. Similarly for Y and Z. Depending in which of the two possible locations for each of X, Y and Z you choose, the triangle XYZ will have different sidelengths.

EDIT: Given the additional assumption that it is known on which side of the plane ABC the points X, Y, Z lie, respectively, they are determined uniquely (see comment). The OP requested an example for $|AB|=1149, |BC|=1730, |CA|=1016, |AX|=1054, |AY|=1872, |AZ|=1914$, $|BX|=1818, |BY|=2445, |BZ|=2163, |CX|=102, |CY|=856, |CZ|=1020$. Assume further that X, Y and Z lie all in the same half space determined by the plane ABC. We can proceed as follows: Set $A=(0,0,0)$ and $B=(1149, 0, 0)$. C can be chosen as the intersection of the circles with radii 1016 and 1730 around A and B, respectively, in the xy-plane. I.e., in order to find a suitable C, we solve

$$x^2+y^2=1016^2, \hspace{0.5cm} (x-1149)^2+y^2=1730^2$$

WLOG we choose the solution with $y>0$ and obtain $x=-\frac{213481}{766}$ and $y=\frac{15\sqrt{2489370063}}{766}$. We can obtain X by finding the intersection of the spheres (a) around A with radius 1054, (b) around B with radius 1818, (c) around C with radius 102, that has positive z-coordinate. That is, we need to solve

$$x^2+y^2+z^2=1054^2, \hspace{0.5cm} (x-1149)^2+y^2+z^2=1818^2, \hspace{0.5cm} \left(x+\frac{213481}{766}\right)^2+\left(y-\frac{15\sqrt{2489370063}}{766}\right)^2+z^2=102^2$$

yielding $x=-\frac{874007}{2298}, y=\frac{338107548509}{6894\sqrt{2489370063}}, z=\frac{2}{9}\sqrt{\frac{1938961458551}{2489370063}}$. Solving similar systems, one may also find the coordinates of Y and Z (always choosing the solution with positive z-coordinate as X, Y and Z are supposed to be located in the same half space with respect to the plane ABC), from which one can then easily deduce the lengths $|XY|, |YZ|$ and $|ZX|$.