TITLE: Can a discrete set of the plane of uniform density intersect all large triangles?
QUESTION [43 upvotes]: Let S be a discrete subset of the Euclidean plane such that the number of points in a large disc is approximately equal to the area of the disc. Does the complement of S necessarily contain triangles of arbitrarily large area?
(Added in July 2021:) I have recently learned that this is Danzer's problem. A Danzer set
(https://en.wikipedia.org/wiki/Danzer_set) is a counterexample. It is apparently not so well-known although the beginning of Gowers answer suggests that he has heard of it.

REPLY [3 votes]: I believe that quite a lot of research has been put into related questions, with much of the attention given to the unit square (and even more, the unit cube in higher dimensions).  Some key references are: 
[1] H.Niederreiter, Random Number Generation and Quasi-Monte Carlo Methods (SIAM, 1992) 
[2] J.Beck and W.Chen, Irregularities of distribution (CUP, 1987) 
[3] B.Chazelle, The Discrepancy Method: Randomness and Complexity (CUP, 2000). 
All I can do is quote you various results from these.  (Some of this is covered in Wikipedia-discrepancy.)  The results are typically expressed in terms of "discrepancy": for N points in the unit square, the discrepancy DN can be defined as supJ |A(J,N) - N λ(J)|where J is all subintervals of rectangles aligned with the axes, A(J,N) is the counting function (that is, the number of points inside J), and λ(J) is the Lebesgue measure.  This definition has an extra factor of N compared to [1], but I think this is useful when you scale the unit square to area N: a rectangle of area λ then "should" contain λ points; and the discrepancy is the difference between that and the number it actually contains.
There are results for the best possible behaviour of DN as a function of N.  For the axis-oriented rectangles, it is Ω(log N), and this can be achieved by the Halton-Hammersley sequence.  When J is changed to rectangles with any rotation, discrepancy is Ω(N1/4), and O(N1/4 √log N) can be achieved by jittered sampling.  When J is all convex subsets of the unit square, you've moved to "isotropic discrepancy" JN -- harder to study, but DN ≤ JN ≤ 8√(N DN).  Also, JN is Ω(N1/4).All of this is related to discrepancy of shapes that can contain some points, and the question is whether they contain the "correct" number.  If you start with a shape j1 with n1 points, and add an empty region (while staying in the set J), you can show that the empty region must have area O(2DN), but I can't see a guarantee that an empty shape obeys the Ω limit.
I'm not sure how close this is to an answer.  I certainly agree that it's a nice area, and I recommend those references -- they can all be read, if not fully understood, by someone with limited mathematical education, like me.