Maximal distance of $2d+1$ points on a sphere If one arranges $2d$ points on the sphere $\mathbf S^{d-1}\subset\Bbb R^d$ as the vertices of the regular octahedron, then one can achieve a minimal spherical distance of $\pi/2$ between any two points, and this is best possible.
What if I want to arrange $2d+1$ points on $\mathbf S^{d-1}$ as far apart as possible from each other? What are the best known upper and lower bounds on the minimal distance between any two points in such an arrangement, and is there anything known about how these arrangements look like?
 A: OK, it's late and I may be wrong but I think that you can obtain the $2d$ points by using any set of orthonormal basis vectors $\{v_1,\ldots,v_d\}$ and their negatives.
Now if $n$ is such that a Hadamard matrix exists, you could take the rows of $H$ an $n\times n$ Hadamard matrix in its $\pm 1$ formulation, and its negative. If you then remove the first column, that would leave you with pairwise inner product $\pm \frac{1}{n} \sqrt{\frac{n}{n-1}}$ or something of this ilk. You can then convert this to the angle $\theta.$ This means that your $d=n-1.$
Maybe more general constructions exist, and feel free to shoot down my answer.
A: Turns out kodlu's idea works in all dimensions, regardless of the existence of any Hadamard matrices.
Consider all coordinate permutations of
$$(1,...,1,-d)\in\Bbb R^{d+1}\quad\text{and}\quad (-1,...,-1,d)\in\Bbb R^{d+1}.$$
This gives $2d+2$ points in $\Bbb R^{d+1}$, but all these lie in the $d$-dimensional subspace orthogonal to $(1,...,1)$. 
The smallest angle between two of these is attained e.g. between $a=(1,...,1,-d)$ and $b=(-1,...,-1,d,-1)$, whose cosine turns out to be
$$\cos\measuredangle(a,b)=\frac{\langle a,b\rangle}{\|a\|\|b\|} = \frac{\overbrace{(-1)+\cdots+(-1)}^{d-1}+d+d}{\underbrace{1+\cdots+1}_d+d^2} = \frac{d+1}{d(d+1)}=\frac1d.$$
