I have the following statements for an MILP:
$c$ (can be $1$ or $0$);
$\alpha_j$ (real numbers with $0\le\alpha_j\le1$).
I have a linear inequality system for $\alpha_j$:
$\sum_jv_j\cdot\alpha_j = 0$ (with $v_j$ constants)
$\sum\alpha_j = c$
and I have the following logic:
If there exists a solution for $c=1$, the formulation should be infeasible;
If there exists the only one solution $c = 0$ (each $\alpha_j$ must be $0$) the formulation should be feasible.
I need some more equations or changes so that the logic above holds. The background is the following. I want to test if a point (here $(0.0, 0.0, 0.0)$) is inside a polygon. The constants $v_j$ are the vertices of the polygon. The above equations have to be set up for each spatial direction, here we only focus on $x$. If no solution can be found for $c = 1$, the point is outside. For my calculations I have to make sure that the point is outside.
When I use an additional constraint $c = 1$ the MILP finds a solution for $c = 1$ und no solution for $c = 0$. This helps to identify if $c$ can be $1$ but this flips the feasible solution space since the solver breaks when $c = 0$ which should be the feasible one. Adding the constraint $c = 0$ will not help, since it is not enough that $c = 0$ is one potential solution, it must be the only one valid solution.
When I use the objective function
max(c) I can conclude that
IF max(c) = 1 THEN not feasible (or
IF max(c) = 0 THEN feasible). However I don't want to use $c$ in the objective function.
Is there any other possibility to change the formulation so that the logic above holds?