1
$\begingroup$

Given two points $x_1$ and $x_2$ on the sphere, one can find another set of points on the sphere $\{y_1, y_2\}$ such that the product of Euclidean distances to the given points $x_i$ is the same for both points, that is $$d(y_1, x_1) d(y_1, x_2) = d(y_2, x_1) d(y_2, x_2)$$ These points can be found in practice simply by taking a line parallel to the line joining $x_1$ and $x_2$ on the sphere.

How can this be done for four points $\{x_1, x_2, x_3, x_4\}$? I am interested in finding a set of points $\{y_j\}$ (with at most four elements) such that $d(y_j,x_1)d(y_j,x_2)d(y_j,x_3)d(y_j,x_4)$ is the same for all $j$ (but does not have a known value).

This time, if you connect the points $x_i$ they form a simplex, which means you can't really find anything "parallel" to it. I tried to see if finding a triangle that is parallel to one of the faces of the simplex would work, but it does not seem to give three points with an equal product of distances.

How can this be done?

$\endgroup$
1
$\begingroup$

Condition $$d(y_j,x_1)^2d(y_j,x_2)^2d(y_j,x_3)^2d(y_j,x_4)^2=k \ \ (constant)$$ gives a locus that is a surface with implicit equation:

$$\Pi_{k=1}^4 ((x-x_k)^2+(y-y_k)^2+(z-z_k)^2)=k$$

This surface of degree 8, as it intersects the sphere (your assumption: the constant has already been calculated on a set of four points) will, in general, possess an infinite number of points.

It is doubtful that interesting things, and in particular a geometric contruction, can be found concerning this locus.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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