Triangulations of translation surfaces whose edges are shorter than the diameter Let $S$ be a compact translation surface (i.e. a surface endowed with a singular flat metric such that singular points are locally isometric to a cone of angle an integer multiple of $2\pi$, and that parallel transport along any closed curve is the identity) of genus $ \geq 2$. 
My question is : is there always a geodesic triangulation of $S$ with vertices being singular points such that every edge is shorter than the diameter of the surface ? If true, does it still hold when you relax the conditions, just asking the metric to be flat with conical singularities ?
 A: 
No, there need not be such a geometric triangulation.  

Construction:  Consider $A$, a flat annulus, of width $W$ and length $L$.  Here we assume that $W$ is very large and $L$ is very small.  (That is, take a $W$ by $L$ rectangle and glue the long sides.)
Let $\alpha$ and $\beta$ be the components of $\partial A$.  We glue many sub-intervals in $\alpha$ to isometric sub-intervals in $\beta$, via some complicated permutation.  This gives a high genus surface $S$.  The singular points all live in a graph -- namely the image of $\partial A$ after taking the quotient. (In fancier language: $S$ is the suspension of an interval exchange transformation.)
Proof: Note that the diameter of $S$ is at most $(W+L)/2$.  However, any geometric triangulation must have at least two edges crossing $A$ and these edges have length at least $W$.  So we are done. 
$\newcommand{\diam}{\mbox{diam}}$I actually I tried to prove that the answer was "yes" before realizing that there are counterexamples.  One natural way provide a "yes" answer is via the Delaunay triangulation.  This lead me to Theorem 4.4 of the paper "Hausdorff dimension of sets of nonergodic measured foliations" by Masur and Smillie.  They define $d(S)$ to be the maximal distance of a non-singular point $x \in S$ to the set of singularities $\Sigma \subset S$. Note that $d(S) \leq \diam(S)$.  Their Theorem 4.4 says that all of the Delaunay polygons in $S$ can be inscribed in a disk of radius at most $d(S)$.  Thus the length of any Delaunay edge is at most $2 \cdot d(S)$. 
