Let $\Sigma$ be a smooth convex surface in Euclidean 3-space and $\gamma$ be a unit speed minimizing geodesic in $\Sigma$. Assume that for some $a < b < c$, we have $$\gamma'(a)=\gamma'(b)=\gamma'(c).$$

Is it true that $\gamma'$ is constant on one of two intervals $[a,b]$ or $[b,c]$?

**Comments**

- This is a simplification-variation of an other question I heard from Dima Burago.
- It is not hard t construct an example of a minimizing geodesic such that $\gamma'(a)=\gamma'(b)$, but $\gamma'$ is not a constant on $[a,b]$. (See the example below.)
- I would be also interested in the analog for $n$ points.
- This paper: "Total curvature..." by Barany, Kuperberg, Zamfirescu is relevant.
- Recently, in "On the total curvature..." by Nina Lebedeva and me we answered the original question of Dima Burago. The problem above remains open, likely the answer is "no".

**Example.** I will construct a convex polyhedron, but it is easy to smooth.
Consider polyhedron defined by 5 inequlaities:
$$z\ge 0,\ \ |x+\alpha{\cdot}y|\le \alpha\ \ \text{and}\ \ z\pm\beta{\cdot}y\le \beta$$
and look at the minimizing geodesic between points $(0,1-\epsilon, \beta{\cdot}\epsilon)$ and $(0,-(1-\epsilon), \beta{\cdot}\epsilon)$. For appropriately chousen $\alpha$, $\beta$ and $\epsilon$ this minimizing geodesic will pass through the faces in this order $$\{z+\beta{\cdot}y= \beta\},\ \ \{x+\alpha{\cdot}y= \alpha\},\ \ \{z=0\},\ \ \{x+\alpha{\cdot}y= -\alpha\},\ \ \{z-\beta{\cdot}y= \beta\}$$
and it will have the same velocity vector $(1,0,0)$ on both faces $\{z+\beta{\cdot}y= \beta\}$ and $\{z-\beta{\cdot}y= \beta\}$.

butif you and Dmitri can't answer it, it's not likely I can either. Nice question, though. I will have to give it a try. $\endgroup$ – Deane Yang May 2 '11 at 16:32