Let us fix a constant $r>1$. Let $d(x,y)$ denote the distance between points $x,y\in \mathbb R^2$. Suppose we have a discreet subset $X\subset \mathbb R^2$ such that 


1) For any two points $x,x'\in X$ we have $d(x,x')\ge 1$.


2) For any point $y\in \mathbb R^2$ there is $x\in X$ such that $d(x,y)\le r$.


**Question.** Is it true that there exist two constants $R>0$ and $\alpha>0$ (depending on $r$), such that for every set satisfying conditions 1 and 2, one can find a triangulation of $\mathbb R^2$ with vertices in $X$, so that each triangle has diameter $\le R$ and all three angles $\ge \alpha$?



Finding a not too slim triangulation with prescribed vertices on $\mathbb R^2$
Yes, the Delaunay triangulation will have this property. From 2) the circumradius of any triangle will be at most $r$, so diameter will be at least $2r$. Also a too small angle would either imply an edge of length less than $1$ or a circumradius larger than $r$.


