Let $\pi=PG(2,n)$ be a desarguesian projective plane of odd order $n$ and $\mathcal{L}$ be a subcollection of lines of $\pi$. Then a set $B$ of points of $\pi$ is called a blocking set relative to $\mathcal{L}$ if for every line $\ell$ in $\mathcal{L}$, we have $\ell\cap B\neq\emptyset$.
I know that if $\mathcal{L}$ is the set of all lines in $\pi$ then we have a standard blocking set, of size at least $n+1$ (iff the set is a line).
Now let $\mathcal{O}$ be an oval of $\pi$, hence a (maximal) set of $n+1$ points, not 3 colinear. We say that a line $\ell$ is
- external to $\mathcal{O}$ if $\vert \ell\cap\mathcal{O}\vert =0$.
- tangent to $\mathcal{O}$ if $\vert \ell\cap\mathcal{O}\vert = 1$.
- secant to $\mathcal{O}$ if $\vert \ell\cap\mathcal{O}\vert = 2$.
I'm interested by lower bounding the size of a blocking set $S$ relative to the collection of all external lines to $\mathcal{O}$. A simple counting argument yields $$\vert S\vert\geq \frac{n-1}{2}$$
Indeed each point of our blocking set covers at most $n+1$ lines. We have in total $\binom{n}{2}$ external lines, therefore we need at least $\frac{n-1}{2}$ points.
I would like to build a blocking set matching this bound (if possible)
A trivial blocking set relative to the collection of all external lines is given by a line secant to $\mathcal{O}$ minus the two points on the oval, giving a set of size $n-1$. I need to reduce this set by half.
I tried looking at points on a secant line that are on the exterior of $\mathcal{O}$ (i.e. at the intersection of 2 tangent to $\mathcal{O}$), but they cannot block all external lines. I need another idea.
Note that I will then be interested by blocking relative to all secants, but I guess that if I can build the first one, I should be able to look at the secant case.
