How can I find four colors with maximum equal difference?

Question

I need to find four colors, expressed as triple $(r_i, g_i, b_i)$ where $0 \le r_i,g_i,b_i \le 1$, $0 \le i \le 3$. Define color difference as $D_{i,j}=\sqrt{(r_i-r_j)^2+(g_i-g_j)^2+(r_i-r_j)^2}$.

1st requirement: $D_{i,j}=D_{k,n}$, $0 \le i,j,k,n \le 3$, $i \ne j$, $k \ne n$.

2nd requirement: $D_{0,1}$ is maximal.

I can make system of 6 equations for 1st requirement, however from the second requirement I get huge equation of 6th power, which makes me think that solving system is not a good idea.

Getting parametric equations for that colors will be excellent, but I think they won't be easy calculatable, so at least one solution will be ok.

Any ideas how to get some results?

Answer 1

Each color is a point in a unit cube with coordinates $(r,g,b)$. They are all at equidistance from each other, so they form a tetrahedron. So you just need to find the largest tetrahedron in a cube. There are two :

1. (0,0,0) (1,1,0) (1,0,1) (0,1,1)
2. (1,1,1) (0,0,1) (0,1,0) (1,0,0)