In the following [paper](https://www.renyi.hu/~p_erdos/1978-40.pdf) (pages 122-23), Erdős asks if there is a constant $c > 0$ such that every subset $A$ of plane of area more than $c$ contains the vertices of a triangle of unit area.


Is this still open? Has anyone discovered interesting lower bounds for $c$?


As a "motivational" puzzle you can show that if $c = \infty$ then $A$ contains vertices of triangles of all possible areas.


Addendum: Monroe has suggested looking at outer measure: First note that there is a subset of plane of full outer measure avoiding vertices of unit area triangles. Why? Just use transfinite induction to construct a unit area triangle free set X which meets every compact positive area subset of plane. This is doable because at any stage we just need to avoid less than continuum many "heights" corresponding to the base lengths that we have accumulated by that stage. Similarly for category. But what about the following variation?


Question: Let X be a bounded subset of plane. Must there exist a full outer measure subset Y of X that avoids vertices of triangles of unit area? Of course the answer is yes under CH or MA, but does this hold in ZFC?


To illustrate why this could be hard, let me state a related problem of P. Komjath.


Question: Let X be a bounded subset of plane. Must there exist a full outer measure subset Y of X that avoids rational distances? Can we even avoid unit distance?


There are uncountably many variations on these but they all currently seem very hard to me.



A question of Erdős
According to [Daniel R. Mauldin, Some Problems and Ideas of Erdős in Analysis and Geometry. Erdős Centennial 25 (2013): 365-376.](http://link.springer.com/chapter/10.1007/978-3-642-39286-3_13)
this is still open. At least it was still open in 2013.
In this chapter it is also stated that we can assume that $A=A\_1\cup\cdots\cup A\_n$ where the $A\_i$ are compact convex sets, and that the statement is true if $n\leqslant 3$, while $n=4$ is the smallest open case.


