Characterization of a sphere: sub-spheres have two (antipodal) centers? Suppose we are given a Riemannian manifold $M$ (it has dimension $n$, is metrically complete and connected) with the following property: for every point $p\in M$ and for every radius $0\leq R$ there is another point $q(p,R)\neq p$ and radius $R'(p,R)\in \mathbb{R}$ such that
$$S(p,R) = S(q(p,R),R'(p,R))$$
where, provided $Q\geq 0$, $S(a,Q)\subset M$ is the set of points which are a distance $Q$ from $a\in M$ (If $Q<0$, $S(a,Q)$ is agreed to be empty).
Q: Is $M$ isometric to an $n$-dimensional sphere?
(To see why a sphere has the mentioned property, define $q(p,R)$ simply to be the point antipodal to $p$ and define $R'(p,R)=\pi R^*-R$ where $R^*$ is the radius of the sphere)
Be sure to have a look at a, virtually identical, question I asked on mathoverflow
 A: i) For $p$ we have antipodal point $q$ s.t. $$M= B_r(p)\cup B_R(q)$$
and
$(S:=)\ \partial B_r(p)=\partial B_R(q)$.
For $x\in S$, there is a minimizing geodesic of unit speed $c$
(resp. $c_2$) from $p$ to $x$ (resp. from $q$ to $x$).
For small $\varepsilon>0$, note that $c'(r)$ is orthogonal to
$T_{x}\ \partial B_{\varepsilon } \bigg(c(r-\varepsilon )\bigg)$.
And $c_2 '(R)$ is orthogonal to $T_{x}\ \partial B_{\varepsilon }
\bigg(c_2
(R-\varepsilon )\bigg)$.
Here $B_r(p)$ contains $
 B_{\varepsilon } \bigg(c(r-\varepsilon )\bigg)$ and $B_R(q)$
 contains $
B_{\varepsilon } \bigg(c_{2} (R-\varepsilon )\bigg)$.
Hence interiors of $
 B_{\varepsilon } \bigg(c(r-\varepsilon )\bigg)$ and $
B_{\varepsilon } \bigg(c_{qc(r)} (R-\varepsilon )\bigg)$ do not
intersect so that union of $c$ and $c_2$ is smooth.
ii) Cut locus ${\rm Cut}\ (p)$ of $p$ is $\{q\}$ : If $x\in
  {\rm Cut}\ (p)\cap {\rm Int}\ B_R(q)$ s.t. there are at least two
  minimizing geodesics from $p$ to $x$, then geodesics pass through
  $S$. Hence $x=q$.
Hence $\exp_p\ tv,\ t\in [0,r+R]$ is minimizing for all $|v|=1$.
If ${\rm diam}\ M=d(a,b)$, then this implies that $d(a,b)\leq
r+R$.
So any point ant its antipodal point give a diameter so that it is
isometric to a sphere.
