Point with connected geodesic circles on closed convex surface 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 $t\geq 0$?
 A: 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 $v_1$
