5 points conundrum!?

Question

Imagine if we had 2 particles attached to a central point by equal length strings, which repelled each other very strongly. We can see that they would stretch out in complete opposite directions.

If we included another of these particles on a string, the three particles would form an equilateral triangle, such that they repel each other as much as possible.

If we had four, a tetrahedron would be the result. The four repelling particles would form the corners of an imaginary tetrahedron.

However, with five particles it gets a bit tricky! Can anyone suggest a three-dimensional shape with 5 points, such that 5 of these repelling particles repel one another as much as possible?!

Answer 1

Have you tried solving for the locus (i.e. the 10 distance equations) of the  5 points:

(0, 0, 0) and (xi, yi, zi); i = 1,....,4? Or are you (like me) not in the mood for it?

Answer 2

two tetrahdrons stuck together.

the base of each has three particles then on each side of the plane formed by these, place another particle equidistant from the three. I think since everything is symmetrical that this will work fine.

The three particles of the base lie in the equatorial plane of an enclosing sphere and the other two particles are at the N and S poles.

if you work through the forces I think you will see that they balance.

If you look at the particle at the North pole. the component of force in the x-y plane does cancel out leaving a net force only in the direction away from the center of the configuartion (which is the center of the ecnlosing sphere). The same is true for each of the particles in the plane of the equator and you only need show that it is true for one since all are symmetrical. You can easily see that the forces from the other two particles in the plane balance perpendicular to the radius connecting the third point to the center and leave only a net outward force. And the particles at the poles cancel in the N-S direction and leave only a net outward force. Anyway, the forces are easy to work with since you do not need to use actual forces, the angles between them (at least in the plane) are just 120 degrees for each pair. And to take account of the particles at the poles is fairly easy also just think of a plane through these two particles and then through each of the particles in the plane of the equator. The pole particles can only produce a net force in this plane. Well,  work it out a little. It really isn't that difficult to see that the net force on each particle points directly away from the center. These forces are not the same on each particle but for the three particles in the euatorilal plane they are the same and also for the two pole particles. The point about using  just variables  for the forces (such as F, G or whatever) is that only outward directed forces will remain and these will be balanced and keep the configuration stable.