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Problem Description:

The radii $r$ of 4 spheres are $r_1,r_2,r_3,r_4>0$ .

The center points of the spheres are not on a plane.

The spheres can overlap.

The distance to the surface of the sphere is positive in the case a sphere encloses the point. The cost-function is therefore always positive or 0 (in the case all spheres intersect at exactly one point).

Intuition:

There has to be a unique point where the sum of the distances from the point to the surfaces of the 4 spheres is minimal.

Q1: Is there a proof that a unique minimum exists?

Q2: Is the cost-function monotonically decreasing when approaching this point?

Edit:

  1. In the comments it has been concluded, that the function can not be smooth.
  2. In the comments it has been concluded, that in case of equal radii the point is the Fermats point.
Bulbasaur
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    Taking $r=0$ makes the target point the Fermat point of the tetrahedron having the spheres' centers as vertices. This old question asks about that point. Tweaking arguments in the answers might provide insights to the current question. – Blue Oct 25 '23 at 11:32
  • This question is related aswell: https://math.stackexchange.com/questions/2842409/fermat-point-of-a-tetrahedron – Bulbasaur Oct 25 '23 at 13:13
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    @Blue It seems plausible to me that if all the sphere radii are equal (and non-zero), the Fermat point also minimizes the distance to the spheres. But I don't have a rigorous argument for this. – Michael Seifert Oct 25 '23 at 15:01
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    @MichaelSeifert: "It seems plausible [that] the Fermat pt also minimizes the distance to the spheres." ... I agree, mostly. Minimizing $|PA|+|PB|+|PC|+|PD|$ is equivalent to minimizing $(|PA|-r)+(|PB|-r)+(|PC|-r)+(|PD|-r)$, so provided that no sphere encloses the Fermat pt, that pt would be the desired solution; the pt would also work with enclosing spheres, provided that the "distance" from an interior pt to the surface is considered negative. However, if OP intends non-negative distances throughout —can a "cost function" go negative?— then things would seem to get a bit trickier. – Blue Oct 25 '23 at 19:27
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    @Blue: If the "distance" doesn't go negative then the "cost function" wouldn't be smooth when it crosses the sphere boundary; so that's an argument in favor of that interpretation. But I agree that clarity is needed. – Michael Seifert Oct 25 '23 at 19:43
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    @MichaelSeifert: "If the 'distance' doesn't go negative then the 'cost function' wouldn't be smooth when it crosses the sphere boundary" ... Good point! – Blue Oct 25 '23 at 20:24
  • These comments have led me to clarify the question more precisely:
    1. The radius can be different for each sphere.
    2. I am indeed also interested in the case where one or multiple spheres enclose the point.
    3. The distance from the point pt in the case a sphere encloses the point pt is always non-negative.
    4. Is it monotone?
     – Bulbasaur Oct 26 '23 at 08:37

1 Answers1

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There has to be a unique point where the sum of the distances from the point to the surfaces of the 4 spheres is minimal.

I expect there is the following counterexample. Let the centers of the spheres are the vertices of a regular tetrahedron inscribed in the sphere of radius $1$ centered at the origin $O$ and each sphere has radius $3$. If the cost function has a unique minimum at a point $P$ then by the symmetry $P$ should be the origin. The value of the cost function at the origin is $8$. Pick any vertex $V$ of the tetrahedron. Let $Q$ be the intersection point of the ray from $O$ towards $V$ with the sphere of radius $3$ centered at $V$. Then the value of the cost function at $Q$ should be $3\left(\frac{\sqrt{46}}2-3\right)\approx 1.173$.

Here is the graph of a two-dimensional counterpart.

enter image description here

Alex Ravsky
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