While studying some physical problem we stumbled across a geometrical problem, to which we could not find any solution.

Consider a unit sphere in 3D and a collection of $N$ points on its surface. Now you choose a direction to project the sphere to a 2D-plane, which yields a unit circle and points inside it. We then calculate the maximal distance of the points to the center of the circle, which is what we call the "size of the projection" (i.e., the smallest radius $r$ such that all points lie inside the circle with radius $r$).

Now, for a fixed placement of $N$ points, one can find the direction of projection that that minimizes the maximum over the distances, see this figure:

Figure 1, projection

We are now interested in the worst placement of the point, that maximizes this minimal distance.

More formally, if we parameterize the direction of the projection by a unit vector $\vec{n}$, the distance of a point $\vec{p}$ to the center of the projected circle is given by $$ d_{\vec{n}}(\vec{p}) = \sqrt{1-(\vec{p}\cdot\vec{n})^2} = |\vec{p} \times \vec{n}| $$ and we are looking for the quantity $$ B(n):=\max_{\{\vec{p}_i\}_{i=1}^N} \min_{\vec{n}} \max_i d_\vec{n}(\vec{p}_i). $$

Some edge cases (see figure 2): Figure 2, placement for N=1,2

For $N=1$, no matter how you place it, you can always project along its direction to reach distance zero, thus $B(1)=0$.

For $N=2$, the best placement is choosing the two points orthogonal, yielding $B=\frac{1}{\sqrt{2}}$.

If you have an infinite number of points, you can evenly distribute them around the sphere, and there are always points with distance one. Thus, $B(\infty)=1$.

In between, the function is monotonically increasing.

Interestingly, for $N=3$, the best placement is not given by a trihedron, with three pairwise orthogonal points ($0.816$), but by evenly distributing them on the equator ($B=\frac{\sqrt{3}}{2} \approx 0.866$). For $N=4$, however, a tetrahedronal placement ($B=\frac{4}{3\sqrt{2}} \approx 0.943$) is favored over a planar arrangement.

The question now is the following: Is there a closed expression for $B(N)$ for all $N$? If not, is there an upper bound?

Best, Herimon

  • $\begingroup$ If for a given constellation, you consider which of the distances of any 2 points are equal, you'll come up with a finite number of "types". The extremal constellations must have more or less high symmetries (think of isoperimetric inequalities). E.g. for $n=5$ one of the candidates will be a pyramid over a square. So it should be clear that $B(N)$ is algebraic. From there to closed form is of course a much harder question. $\endgroup$ – Wolfgang Dec 6 '18 at 17:35
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    $\begingroup$ Also note that an equivalent formulation is looking for the thinnest cylinder containing all points. So the question is how thick this cylinder must be in the worst case. The axis of the cylinder must go through the center of the sphere, though I don't think this is a real constraint. $\endgroup$ – Wolfgang Dec 6 '18 at 21:22
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    $\begingroup$ This is equivalent to the problem of covering the sphere with "equatorial strips". I doubt very much that an exact solution can be found beyond the first few values of $N$ but the trivial area bound is always at your disposal and it is, probably, not too far from the truth. $\endgroup$ – fedja Dec 6 '18 at 22:09
  • $\begingroup$ Thank you all for you ideas! @fedja: I think you are right, the problems are equivalent. If I did the calculation correctly, the trivial area bound yields $B(N) \leq \sqrt{1-\frac{1}{N^2}}$. $\endgroup$ – Herimon Dec 7 '18 at 13:08

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