There are at least three distinct simple closed quasigeodesics on convex polyhedra [Mat. Sb. (N.S.), 1949, 25(67) :2, 275–306 Quasi-geodesic lines on a convex surface Pogorelov].
Is the same true for straightest geodesics?
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Straightest Geodesics:straightest_preprint.pdf
--Thanks, Bryan
I assume by straightest geodesic you mean: has the same surface angle to either side at every point.
Let the polyhedron $P$ be a doubly covered obtuse triangle $\triangle abc$.
The path $abc$ is a straightest geodesic.
It seems the other quasigeodesics guaranteed by Pogorelov's theorem must pass through vertices, such as that following the
altitude through the obtuse vertex $c$. But this will not be straightest.
The path $aba$ is a quasigeodesic, with less than $\pi$ to each side of every point, but it
is not straightest.
So it seems that this "polyhedron" has just one simple (noncrossing), straightest geodesic.
