Need a hint: the sum of vectors is zero and their norms are equal? Vectors $a$, $b$ and $c$ all have length one. $a + b + c = 0$. Show that 
$$
|a-c| = |a-b| = |b-c|
$$
I am not sure how to get started, as writing out the norms didn't help and there is no way to manipulate 
$$
|a-c| \le |a-b| + |b-c|
$$ 
to get an equality. I just need an idea of where to start.

The norm-square of $a-b$ is
$$
\|a-b\|^2 = \|a\|^2 + \|b\|^2 - 2(a,b) = 2-2(a,b).
$$
Thus the claim is equivalent to prove that
$$
(a,b)=(b,c)=(c,a)
$$
with $(\cdot,\cdot)$ denoting the scalar product. Since $a+b+c=0$, it holds
$$
(a+b+c,a) = (b,a+b+c),
$$
which implies
$$
1 + (b,a)+(c,a) = (b,a) + 1 + (b,c)
$$
hence $(b,a)=(b,c)$ and so on. I think you can complete now.
