# Characterization of a sphere: every “sub-sphere” has two centers

Let me ask this question without too much formalization:

Suppose a smooth surface $M$ has the property that for all spheres $S(p,R)$ (i.e. the set of all points which lie a distance $R\geq 0$ from $p \in M$, with distance as measured inside the surface), there is always a different point $q(p,R) \in M$ and a distance $D(p,R)\geq 0$ so that $$S(p,R)=S(q(p,R),D(p,R))$$ In words: both $p$ and $q(p,R) \neq p$ are a center of the sphere $S(p,R)$.

Q: Is $M$ (a part of) a sphere? More formally: is $M$ isometric to (a subset of) a sphere?

Remark: Spheres clearly possess the mentioned property, with every sub-sphere having one point $p$ ánd its antipodal $\pi(p)$ at its center. This property is maintained when we remove a finite number of pairs of antipodal points from such a sphere or when we take a few disconnected spheres.

This question has a counterpart on stackexchange.

Blaschke's conjecture might be relevant. In particular, if the surface is $S^2$ with a $C^3$ metric, then it follows from Blaschke's conjecture, proved by Green. This states that a metric on the sphere in which every point has a unique conjugate point must be the round metric on the 2-sphere. In your condition, one may observe that $q(p,R)$ does not depend on $R$, essentially by the Gauss lemma. The point is that the sphere of radius $r$ about $B(p,R)$ (the ball of radius $R$ about $p$) is $S(p,r+R)$. Hence $q(p,R)=q(p,r+R)=q(p)$. Then $q(p)$ is the unique point conjugate to $p$.