For every point set $A \subset R^2$, prove that in general the sum of the coordinates of $\phi(T)$ is independent of a triangulation T and is associated to the area of the Convexv_Hull(A).
We define vector $\phi(T) \subset R^n$, where $n$ is the number of points that we use in the triangulation*. Every $i-th$ coordinate of $\phi(T)$ is equal to the sum of areas of the triangles in T that contain $p_i$.
That looks pretty straightforward to me (since no matter how we partition the space, the area is a constant value$), intuitively, but how to approach its proof?
*we assume that a triangulation uses all points of a point set
$As the Cake By The Ocean song says: And no matter how you slice it...it's a piece of cake.
Given any triangulation $T$ of $A$, let $t_1, t_2, \ldots, t_f$ be the triangles of $T$ and $a_1, a_2, \ldots, a_f$ be the corresponding area. Define
$$\theta_{ij} = \begin{cases} 1, & p_i \in t_j\\ 0, & \text{ otherwise } \end{cases} \quad\text{ for }\quad 1 \le i \le n;\; 1 \le j \le f$$ We have $\displaystyle\;\phi(T)_i = \sum_{j=1}^f \theta_{ij} a_j\;$ and $\displaystyle\;\sum_{i=1}^n \theta_{ij} = 3\;$. As a result, $$ \sum_{i=1}^n \phi(T)_i = \sum_{i=1}^n \sum_{j=1}^f \theta_{ij} a_j = \sum_{j=1}^f \sum_{i=1}^n \theta_{ij} a_j = \sum_{j=1}^f 3 a_j = 3\sum_{j=1}^f a_j = 3\,\verb/Area/(\textrm{co}(A))$$