The problem is to prove Helly's theorem for the case, when the convex bodies are polytopes, by using linear programming duality.
Helly's theorem
Let $B_{1},...,B_{m}$ be a collection of convex bodies in $\mathbb{R}^{d}$ with $m>d$. If every $d+1$ of these convex bodies have a point in common, then all $m$ of these convex bodies have a point in common.
Ideas for solution
By definition, polytopes is intersection of the half spaces, so intersection of polytopes is intersection of half spaces, therefore we can define inequalities $Ax. So we get linear programming problem.
In order to prove it, we can take a look at equivalent problem, according to Helly's theorem, $Ax (intersection of half spaces) doesn't have solution, when any $n+1$ selected inequalities don't have solution.
We should state dual LP problem, which should be feasible and unbounded. Next step is to show that $n + 1 $ nonzero dual variables sufficient for an unbounded solution.
Questions
Unfortunately, I can't figure out what exactly the dual problem will represent? And which objective function should be used? In contradiction to this problem, other usage of duality, in proving minimax theorem, was very clear. We have two LP problem, primal and dual (minimize the loss and maximize the payoff), where objective functions are loss and payoff.
Let's elaborate a little bit
It's primal LP problem
$max \space 0$
$Ax
It's Dual
$min \space b^Ty$
$A^Ty=0$
$y \geq 0$
So now, the problem is prove that the dual LP problem is unbounded using $n+1$ nonzero dual variables.
But constraints in the dual problem is the Homogeneous systems. What we know about homogeneous system,
1) if $u$ and $v$ are two vectors representing solutions to a homogeneous system, then the vector sum $u + v$ is also a solution to the system.
2) If $u$ is a vector representing a solution to a homogeneous system, and $r$ is any scalar, then $ru$ is also a solution to the system.
3) The solution set to a homogeneous system is the same as the null space of the corresponding matrix A.
So may be using $n+1$ feasible solution we can show that there is infinite many basic feasible solutions, therefore the problem can be unbounded. I am sure this doesn't make sense. But I don't have a better explanation.
The second idea, actually by strong duality theorem, if the primal problem is infeasible that the dual problem is unbounded, but in this case, how can I apply $n+1$ nonzero solutions?