# 5 Points uniformly placed on a sphere

I have a sphere and I have to place some points on it, the most uniformly possible.

If I have 4 points, placing them as vertices of a tetrahedron seems good. If I have 6 points, placing them as vertices of a octahedron seems very good too.

How can I find a way (the best if it exists) to place only 5 points ?

EDIT : "Uniformly" would mean that if I draw a Voronoï diagram on the sphere, each point has a same-area cell and the diameter of a cell is minimized (they are "round" and not some thin slices of the sphere).

• The notion of 'best' seems ambiguous, but it seems that three points on the equator, and the other two points at each pole seems a good candidate for your problem. – Sangchul Lee Aug 22 '12 at 13:59
• This might be useful www2.research.att.com/~njas/packings – leonbloy Aug 22 '12 at 14:01
• Searching for "distributing points on a sphere" turns up many things. It is a hard problem, with the usual definition being to maximize the minimum distance between the points. I believe sos440's solution has been proven optimal in the 5 point case, but higher cases are difficult to prove. – Ross Millikan Aug 22 '12 at 14:05
• A triangular dipyramid has 5 vertices. You might have to "squash" it a bit to get the points onto a sphere, but similar to the octahedron for 6 points you'll get a point at each "pole" and the rest distributed around the "equator". – KeithS Aug 22 '12 at 14:46
• I'd go with @Keith's and sos440's take. The VSEPR theory in chemistry maintains that molecules that consist of one central atom with five pendant atoms (e.g. phosphorus pentachloride, $\require{mhchem}\cf{PCl5}$) take on a trigonal bipyramidal arrangement, like this. – J. M. is a poor mathematician Aug 22 '12 at 15:10

The answer in this case for $n=5$ would be:
More answers for other values of $n$ can be found here.