Point with connected geodesic circles on closed convex surface

Question

Let $S\subset\mathbb{R}^3$ be a closed convex surface (smooth if necessary) and denote by $d$ the intrinsic distance on $S$. Can we always find a point $p$ such that $S_p(t):=\{q\in S\mid d(p,q)=t\}$ stays connected for all $t\geq 0$?

Answer 1

Consider a 2-dimensional flat equilateral triangle $T_i$. If we glue along boundary $\partial T_1= \partial T_2$, then we have a convex surface $\Sigma$. Then note that any point does not hold the property

Notation and setting : If $v_i$ are three vertices in $\Sigma$, then fix an interior point $x\in T_1$. Assume that $v_1$ is the closest vertex to a point $x$. And we define $y_i\in [v_1v_i] $ s.t. $$ |v_1y_2|=|v_1y_3|=\varepsilon$$ and $v_1y_2xy_3$ is in circle of radius $\frac{1}{\sqrt{3}}\varepsilon$ and center $o$

Now we will show that some geodesic sphere of $x$ is not connected :

(1) If $f(\theta )=|xy_2|+|xy_3|+|y_2y_3|$ where $\theta=\angle(\overrightarrow{ oy_2},\overrightarrow{ox} )$ then $f$ has a maximum at $ \theta = \frac{\pi}{3}$. And $$f(\frac{\pi}{3} ) =r,\ r=\frac{2+\sqrt{3} }{\sqrt{3} } \varepsilon$$

Here $r<r_0$ where $r_0= \frac{4}{\sqrt{3}} \varepsilon = 2|v_1x| $

(2) Assume that a point $x_2\in T_2$ satisfies that there are at least two shortest paths from $x$ to $x_2$. Further note that there is a point $x_2$ s.t. $|xx_2|$ has the minimum. Hence $|xx_2|\leq r$. Hence for some $R\leq \frac{r+r_0}{2}$, the geodesic sphere $S_R(x)$ is not connected, since the geodesic ball $B_R(x)$ does not contains a vertex $v_1$