Let $\mathcal{F} = \{P_1, P_2,\ldots,P_m\}$ be a family of (closed) convex polygons in the plane, each represented by their vertices in (say) clockwise order. Let $n$ be the total number of vertices (you can assume general position, no two vertices are the same, no edges intersects a vertex other than its endpoints, no three edges intersect at the same point).
I'm curious to know if you can you compute a subfamily of largest cardinality possible such that all polygons share a single common intersection point in $\mathcal{O}(n^2)$ time? (In the Real RAM model, i.e. you can store an arbitrary precision real number in a register and perform arithmetic operations on them in constant time)
The best I've managed to come up with is $\mathcal{O}(n^2\log(n))$:
- I can modify the Bentley-Ottman algorithm (with the segments being the polygon edges) so that at every event point $p$, I count the number of "lower hull" edges and "upper hull" edges under $p$ on the sweep line (modifying my search tree so this can be done in $\mathcal{O}(\log(n))$ time) and compute the difference, giving me the number of polygons containing $p$. I simply keep track of the "best" point $p$, then test every polygon against $p$ in $\mathcal{O}(n)$ total time.
Alternatively, by computing the $\mathcal{O}(m^2)$ points which are the leftmost in the intersection of a pair of polygons (which can be shown to be doable in $\mathcal{O}(nm)$ time) and then counting the total number of polygons which contains each point (in $\mathcal{O}(nm^2)$ time), you can solve the problem in $\mathcal{O}(nm^2)$ time, which is better for small $m$ (and you can probably also get $\mathcal{O}(n^\alpha m^\beta)$ for different values of $\alpha + \beta > 2$ using more sophisticated simplex range counting datastructures).
Is anything faster possible (asymptotically)? Or is there some (conditional) lower bound showing that $\mathcal{O}(n^2)$ is hard/impossible to achieve?
