A generalization of the Borsuk-Ulam theorem? In the setting of the Borsuk-Ulam theorem, is it true that for any distance $d$ there exist two points $A$ and $B$ with distance $d$, such that $f(A)=f(B)$? (Two antipodal points having the maximal distance).
 A: Remark:  As pointed out by Thomas Anton, the arguments below do not make sense: $f(u(t))$ is an element of $\mathbb R^n$, not an element of $S^n$. Thus $B = f(u(t_r)) \notin S^n$. And even if $B$ were in $S^n$, there is no reason to expect that $f(A) = f(B)$.
I tried to delete my answer, but this is impossible for accepted answers.
Here is the original answer:
Yes. The Borsuk-Ulam theorem states that for any continuous map $f : S^n \to \mathbb R^n$ there exists $x \in S^n$ such that $f(x) = f(-x)$. The points $x, y = -x$ have the distance $\lVert y - x  \rVert = 2$. Now let $u : [0,1] \to S^n$ be any path from $x$ to $y$. The function $\phi : [0,1] \to \mathbb R, \phi(t) = \lVert f(u(t)) - x \rVert$, is continuous and satisfies $\phi(0) = 0, \phi(1) = 2$. By the intermediate value theorem for each $r \in [0,2]$ there exists $t_r \in [0,1]$ such that $\phi(t_r) = r$. Now take $A = x, B = f(u(t_r))$.
By the way, this proof works for any metric $d$ on $S^n$ which generates the standard topology.
