Conclusion

The question was small: how few points can contain $r$ lines in every direction of a finite plane? The answer turned out to have three parts of quite different character.

The first part is an identity. The size of a union of $r$ lines per direction exceeds $\frac r{r+1}q(q+1)$ by exactly the variance of its multiplicities about $r+1$, so the union is smallest when the multiplicities are constant, and then it is a dual maximal arc. Everything about the equality case follows from the existence theory of maximal arcs: attained for even $q$ whenever $r+1$ divides $q$, never attained for odd $q$ in the range $1\le r\le q-2$. The Ball–Blokhuis–Mazzocca theorem that decides the odd case is a polynomial-method theorem, while the direct polynomial method, as Chapter 7 shows, cannot see past $r=1$. I find this the most instructive feature of the problem: the method's power over finite fields comes from polynomials of degree below $q$, and an $r$-fold union is the zero set of nothing of that degree, yet the method reappears once the problem is dualised, in the guise of a nonexistence theorem about polynomials of degree far above $q$.

The second part is exact and elementary. Near $r=q$ the problem is about the points that all of the omitted lines pass through, and it is solved by the observation that $s+1$ points can be avoided by $q-s$ lines in every direction only if they determine every direction. The threshold at which the naive answer $q^2-s$ fails is the least size $D(q)$ of a set determining all directions, a quantity that seems not to have been studied and that I could compute only for $q\le16$, where it equals the obvious counting bound except at $q=9$.

The third part is open. For odd $q$ and fixed $r\ge2$ the deficiency $\kappa_r(q)-\frac r{r+1}q(q+1)$ is positive, but I cannot show that it grows with $q$, let alone linearly as it does for $r=1$. The constructions I have, from pencils of conics, are optimal at $q=7$ and far from optimal asymptotically. The data suggest that extremal configurations have multiplicities within two of $r+1$; if that could be proved, the deficiency question would become a counting problem about sets of type $(r-1,r,r+1,r+2)$ with respect to a pencil.

Open problems.

1. For odd $q$ and fixed $r\ge2$, determine the order of growth of $\Delta_r(q)=\kappa_r(q)-\frac r{r+1}q(q+1)$ (Question 6.3). The first case is $\kappa_2(q)$ for odd $q$; the first unknown value is $\kappa_2(9)\in[61,65]$.
2. For $q=2^h$ and $(r+1)\nmid q$, determine $\kappa_r(q)$; the first cases are $\kappa_2(2^h)$, known only for $h\le3$ ($14$ and $51$), and $\kappa_4(8)\in\{58,59\}$.
3. Determine $D(q)$ in general. Is $D(q)=t_0(q)=\lceil(1+\sqrt{8q+9})/2\rceil$ for all $q\ne9$? What is special about $AG(2,9)$, in which no five points have ten pairwise non-parallel joining lines?
4. Determine $\tau_s(q)$ for $D(q)-1\le s\le q-2$, equivalently $\kappa_r(q)$ for $2\le r\le q+1-D(q)$; the data for $q=7$ ($\tau_4=5$, $\tau_5=9$) and $q=8$ ($\tau_5=10$, $\tau_6=13$) show that $\tau_s$ grows quickly once it leaves $s$.
5. Classify the extremal $2$-fold configurations for odd $q$, as Blokhuis and Mazzocca did for $r=1$. For $q=7$ there are at least two inequivalent ones.
6. Extend to $AG(n,q)$, $n\ge3$, where lines in different directions need not meet and the counting identity of Chapter 3 has no direct analogue; the relation to the Furstenberg-set problem of Ellenberg and Erman (2016) and Dhar, Dvir, and Lund (2021), which also concerns sets containing many points of many lines, should be clarified.

The companion dissertation argued that the polynomial method rests on three lemmas and that its applications differ in the bridge between them. The present problem shows a fourth thing: sometimes the bridge does not exist, the method gives only its trivial bound, and the answer comes from elsewhere. But "elsewhere" was itself built with polynomials.