How can points be distributed on a Unit Sphere such that they maximize the minimum distance between any pair of
points? In 1943, Fejes Tóth proved that for points, there always exist two points whose
distance is

and that the limit is exact for , 4, 6, and 12.

For two points, the points should be at opposite ends of a Diameter. For
four points, they should be placed at the Vertices of an inscribed Tetrahedron.
There is no best solution for five points since the distance cannot be reduced below that for six points. For six points,
they should be placed at the Vertices of an inscribed Octahedron. For seven points,
the best solution is four equilateral spherical triangles with angles of 80°. For eight points, the best dispersal
is *not* the Vertices of the inscribed Cube, but of a square Antiprism
with equal Edges. The solution for nine points is eight equilateral spherical triangles with
angles of
. For 12 points, the solution is an inscribed Icosahedron.

The general problem has not been solved.

**References**

Ogilvy, C. S. *Excursions in Mathematics.* New York: Dover, p. 99, 1994.

Ogilvy, C. S. Solved by L. Moser. ``Minimal Configuration of Five Points on a Sphere.'' Problem E946.
*Amer. Math. Monthly* **58**, 592, 1951.

Schütte, K. and van der Waerden, B. L. ``Auf welcher Kügel haben 5, 6, 7, 8 oder 9 Pünkte mit Mindestabstand Eins Platz?''
*Math. Ann.* **123**, 96-124, 1951.

Whyte, L. L. ``Unique Arrangement of Points on a Sphere.'' *Amer. Math. Monthly* **59**, 606-611, 1952.

© 1996-9

1999-05-26