# Projective plane - Blocking set relative to line from an oval

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.

• either 2 tangent lines, and therefore $$\frac{n-1}{2}$$ secant lines and $$\frac{n-1}{2}$$ external lines.
• or 0 tangent line, and therefore $$\frac{n+1}{2}$$ secant lines and $$\frac{n+1}{2}$$ external lines.
Therefore each point of our blocking set covers at most $$\frac{n+1}{2}$$ lines and we have $$\vert S\vert\geq n-1$$ Then my construction is optimal.