Finding a not too slim triangulation with prescribed vertices on $\mathbb R^2$

Question

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 $\ge \alpha$?

Answer 1

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 $r$.

Answer 2

The Delaunay triangulation produces a triangulation with the desired the guarantees. 2) ensures that the circumradius of any triangle in the triangulation is at most $r$. A circle of radius $r$ containing the triangle means the diameter of any triangle is at most $2r$, i.e., the diameter of any triangle is at most $R = 2r$.

1) ensures the length of any edge of any triangle is at least $1$. For any Delaunay triangle $T$ with circumradius $r_T$, the law of sines can be applied to any angle $A_T$ and opposite edge $a_T$,

$\sin A_T = \frac{a_T}{2r_T} \leq \frac{1}{2r}.$

So all angles of the triangulation are at least $\alpha = \arcsin\left(\frac{1}{2r}\right)$.

Note that while a unique Delaunay triangulation is only guaranteed to exist if the pointset is in general position (i.e., no four parts are cocircular), these guarantees hold for any set of input points by using the Delaunay construction and picking any arbitrary triangulation for any sets of four of more cocircular points.