Following this post, I have been thinking about the problem posed by Erdős,

Does there exist a constant $c > 0$ such that every subset $A$ of the plane of area more than $c$ contains the vertices of a triangle of unit area?

I think the spirit of the problem justifies assuming $A$ is measurable (counterexamples exist for non-measurable sets, see post), and so we can use the following to simplify:

If $A$ is measurable, with

finitemeasure $a$, then for every $\epsilon >0$ there exists an $n \in \mathbb{N},$ $r > 0,$ such that $A$ can be filled with $n$ non-overlapping disks of radius $r$ with combined area $a' = n \pi r^2 > a(\eta_h -\epsilon)$, where $\eta_h \approx 0.9$ is the optimal packing density of disks in the plane.

So we can reduce to just considering sets of nonoverlapping disks of equal radius $r$. The problem becomes,

Does there exist a constant $c > 0$ such that every set $A$ of $n$ disks of radius $r$, with $n \pi r^2 > c$, contains the vertices of a triangle of unit area?

Now, we can scale such a set $A$ by $1/r$ (that is, probably expanding it, because probably $r < 1$.) and ask of the new set of $n$ *unit* disks, if $n\pi r^2>c$, for some $c$, must $A$ contain the vertices of a triangle of area $1/r^2$? Defining $\ell = c/{(\pi r^2)}$, if $\ell<n$, must $A$ contain the vertices of a triangle of area $\pi \ell/c$? That is, must $A$ contain vertices of triangles of all areas $< \pi n/c$? So, setting $c' = c/\pi$ (to get rid of the constant), we get the equivalent question,

Does there exist a constant $c' > 0$ such that every set of $n$ unit disks in the plane must contain the vertices of triangles of all areas $< n/c'$?

Say we are given some such $c'$, and we must determine the conditions under which a triangle of area $c < c'$ must have vertices in $A$. Now, since two unit disks with centers a distance $> n/c$ apart always contain the vertices of a triangle of area $n/c$, we can assume that these disks are all within a distance of $n/c$ of each other. Thus, they all lie in a big disk $D$ of radius $\frac{2n}{\sqrt{3}c}+2$, and area $d_c$.

We can then look at *pairs of points* within this disk that must not *both* lie in $A$. That is, pairs of points in $D$ which, together with some point in $A$, form the vertices of a triangle of area $n/c$. We denote this set $\mathcal{F}_2$.

Using the usual product measure, we denote the measure of $\mathcal{F}_2 \in D \times D$, $m_2$. The measure of $D \times D$ itself is $d_c^2$. We can show that,

$m_2$ = $\Omega(cd_c^2/\sum_{1 \leq i \leq j \leq n} {s_{ij}})$

as a function of $c$ and $n$, where $s_{ij}$ is the distance between the centers of the $i^\text{th}$ and $j^\text{th}$ disks.

Letting $S = \sum_{1 \leq i \leq j \leq n} {s_{ij}}$, we win if $S = o(c)$ (S a function of both $c$ and $n$).

Now, we can also look at the set of *individual* points in $D$ that are "forbidden," i.e., together with two points from $A$, form a triangle of area $n/c$. We denote this set $\mathcal{F}_1$, and take it as a subset of $D$, with measure denoted $m_1$.

It can be shown that the set of points $f_{ij}$ that are forbidden due to forming a triangle of area $n/c$ with points in disks $i$ and $j$, which we say has measure $m_{1,ij}$, satisfies,

$m_{1,ij} = \Omega(d_c/s_{ij})$

$\mathcal{F}_1$ is the union of the $f_{ij}$. Naïvely summing the $m_{1,ij}$, we get the sum $m_1'$ of their measures is,

$m_1' = \Omega(d_cS)$.

If this were indeed $\mathcal{F}_2$, we would win if $S \neq o(c)$, since $d_c \sim \frac{2n}{\sqrt{3}c}$ (winning meaning $m_1 > d_c$, a contradiction). Since, from before, if $S = o(c)$ we also win, the problem would be finished. You could simply pick $c$ large enough, and either $\mathcal{F}_1$ or $\mathcal{F}_2$ would exceed the measure of its superset, a contradiction.

But of course the area of a union is not the sum of the areas. And I can't find a way to deal with the intersections of the $m_{1,ij}$. I also can't tell if I'm really right next to a solution, or if I'm just hitting the brick wall surrounding the problem in a different spot.

I hope you found this interesting, at least. If the way I put growth rates is too confusing, I'll post some updates to clarify. If you have any ideas, please let me know.