Consider a point set $P\subset \mathbb{R}^2$. Let $T$ be a triangulation of $P$. For $t\in T$ a triangle, define $l(t)$ as the length of the longest side of the triangle. I want to find a triangulation that minimizes $\sum_{t\in T}l(t)$, that is the sum of the longest lengths of each triangle in $T$. Is there a name for this particular triangulation?
Note that this is NOT the minimum weight triangulation (https://en.wikipedia.org/wiki/Minimum-weight_triangulation) , nor is it equivalent to the minimum weight triangulation. Here is a counter example below.
Example: Consider the five point set $\{(0,0), (0,2), (2,0), (2,2), (1,-\epsilon)\}$ where $\epsilon$ is an arbitrarily small positive number. Consider the following two triangulations.
Denote $T_1$ by the triangulation with triangles $\{\{(0,0),(0,2),(2,2)\}, \{(0,0),(2,0), (2,2)\}, \{(0,0), (1,-\epsilon), (2,0)\} \}$
Denote $T_2$ by the triangulation $\{\{(0,0),(0,2),(1,-\epsilon)\}, \{(0,2),(1,-\epsilon), (2,2)\}, \{(2,2), (1,-\epsilon), (2,0)\} \}$
Let $\epsilon = 1/4$
The weight (sum of perimeters of all triangles) for $T_1$ is approximately 17.718 The weight (sum of perimeters of all triangles) for $T_2$ is approximately 17.91
$\sum_{t\in T_1}l(t)=7.657$
$\sum_{t\in T_2}l(t)=7.387$
Note that $T_1$ has the lower weight, but $\sum_{t\in T_1}l(t) > \sum_{t\in T_2}l(t)$.