Dispersion of points on a 4D sphere In ℝ4 surface S is defined as defined as all points 1 unit from the origin. Taking the 5 points P1, P2... P5, that all lie on the surface S, move the points along S such that the distance between every point and every other point is maximised. Calculate the angle made between any 2 points and the origin Pn→O→Pk (Where n and k are two different natural numbers from 1 to 5 and O is the origin ). 
I came up with this problem and my approach to solving it was as follows: First I took the simpler version of the problem of the problem in ℝ2 where the surface S is defined the same way but only 3 points, P1, P2, P3. I assumed that these 3 points would form an equilateral triangle and solved for the angle between the points and the origin which was 2π/3. Then I took it up a dimension into ℝ3 with 4 points, I again assumed that the 4 points would form a tetrahedron and used the points P1, P2, P3 and P4 and the origin to find the angle. So to follow the pattern forward into ℝ4 I looked for a point equidistant to all 4 points of a tetrahedron to make the 4-d version of the equilateral triangle then I used geometry to find the answer which was roughly ~1.64 radians.
An additional note, the angle between any 2 of 3 dispersed points on the surface S in ℝ2 is 2π/3. The angle between any 2 of 3 points on surface S in ℝ3 should also be 2π/3 because in any dimensional space 3 non collinear points is cocyclic, I do not have a proof for this either I simply deduced it by intuition. 
Although I'm pretty sure my answer is correct its far from rigorous and when I talked to a mathematician he agreed with the logic of my method but said if I were to prove my result I would have to create some function that represented that problem and maximise that function. My question was, what would that function be and how would you rigorously prove the solution to this problem. 
 A: In the case of $S^1 \in \Bbb R^2$ (three points on the unit circle), you might as well assume that one of them is at the point $(1, 0)$; the others are then at points $(x, y)$ and $(u, v)$, and you're trying to maximize
$$
h(x, y, u, v) = (x-1)^2 + y^2 + (u-1)^2 + v^2 + (x-u)^2 + (y-v)^2,
$$
all subject to $x^2 + y^2 = 1$ and $u^2 + v^2 = 1$. That's a Lagrange multiplier problem waiting to be done.
Why is $h$ the right function? Well... it's the sum of the squares of the distances, rather than the sum of the distances, and there's some proving to do if you're going to believe that's the right thing to look at. But the computation is relatively easy. We compute the gradient of $h$:
$$
\nabla h(x,y,u,v) = 
2\pmatrix{
(x-1) + (x-u)\\
y + (y-v)\\
(u-1) - (x-u)\\
v - (y-v)} = 2\pmatrix{
2x-(1 +u)\\
2y -v\\
2u-(x+1)\\
2v - y} 
$$
We have to make this equal to a linear combination of the gradients of the two constraint functions, which are $$
2\pmatrix{x\\y\\0\\0}, 2 \pmatrix{0\\0\\u\\v}.
$$
so we need to find coefficients $c$ and $d$ such that
$$
2x - u - 1 = cx\\
2y - v = cy\\
2u - x - 1 = du\\
2v - y = dv
$$
while also $u^2 + v^2 = x^2 + y^2 = 1$. To simplify things, I'm going to replace $c$ and $d$ by $2c$ and $2d$, to get 
$$
2x - u - 1 = 2cx\\
2y - v = 2cy\\
2u - x - 1 = 2du\\
2v - y = 2dv\\
$$
so that
$$
(2-c)x - u = 1\\
(2-c)y - v = 0\\
(2-d)u - x = 1\\
(2-d)v - y = 0\\
$$
Now you have to solve this equation, along with $u^2 + v^2 = x^2 + y^2 = 1$, to get the resulting coordinates. It'll turn out to be easiest (I think!) to look at the second and 4th equations and realize that $v/y$ is both $2-c$ and $\frac{1}{2-d}$, so $2-c$ and $2-d$ are reciprocals; that lets you eliminate $d$ from the equations. I think that the first and third then let you solve for $c$. And then it's a matter of taking those numbers, and two two quadratics, and seeing where they lead you. 
Of course, for this problem, it'd be a lot easier to work with the angular coordinate $\theta$; you want to minimize $\theta_1^2 + \theta_2^2 + (\theta_2 - \theta_1)^2$ (where those are the counterclockwise angles from $(1,0)$ to the second and third points. This is a 1D optimization, and easy to solve. 
IF you knew a nice parameterization of $S^3 \subset \Bbb R^4$, you could do the same thing. So-called "euler angles" work OK, although they have singularities that make things messy, and you'll end up with a lot of sines and cosines. 
A possibly better approach is to make some sort of symmetry argument: say that if you have 5 points anywhere on the sphere, and then "symmetrize" them by some sort of averaging over all symmetries of the 3-simplex, the resulting new set of  points has a better (or at least "as good") "sum of distances"; this proves that the points invariant under this symmetry must be optimal. Of course, you need to define an "averaging" operation, and prove its "improves the score" property, and neither of these is easy to do (and this approach doesn't work for less "natural" numbers of points on a $k$-sphere). 
IN short: it's kind of a mess. 
A: Can't you use symmetry argument ? You're looking for a pentachoron (also known as 5-cell or 4-simplex).
