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?

1$\begingroup$ Can you make precise the senses of these approximations? Do you mean disks of increasing size with any fixed center? Do you mean triangles of large area? $\endgroup$ – Andrew Critch Oct 29 '09 at 18:51

$\begingroup$ Fix a point $P$ of the plane. The density of $S$ is then asymptotically uniform if the limit $\sharp\{\box{number of points of }S\mathbox{ in a closed disc centered at P with radius R}\}/R^2$ is $\pi$ (or any nonzero constant, it does not matter) for $R\rightarrow \infty$. Yes, I mean triangles of large areas. $\endgroup$ – Roland Bacher Oct 29 '09 at 19:16

6$\begingroup$ Can I ask how much you care about this, and how much research you've put in? This problem is really addictive, and I don't want to put too much thought into a problem that might have already been solved by some Hungarian twenty years ago. $\endgroup$ – David E Speyer Oct 30 '09 at 0:16

$\begingroup$ See Roland's answer below, which includes a reference to previous work. I'm tagging this "openproblem"! $\endgroup$ – Scott Morrison♦ Nov 1 '09 at 0:34
Edit: The answer below is somewhat obsolete, due to the recent paper of Noga Alon, where he establishes the first nontrivial lower bounds for strong epsilonnets.
Old answer: This answer uses some terminology from discrete geometry. The definitions, and more background material can be found in, for example, "Discrete geometry" by Jiří Matoušek chapters 9 and 10. Less thorough introduction can be consulted without going to the library.
If one drops the requirement that set has to have asymptotic density, and merely asks for the set to have a constant density on some large (but fixed) ball, and restricts the triangles to live inside that ball, the problem does not become much easier. In that case problem asks for a construction of an 1/rnet of size O(r) for the family of all triangles with respect to the Lebesgue measure on [0,1]^2 (or equivalently for a construction of 1/rnet for a grid {1,..,N}x{1,...,N} for large N). The upper bound of r*log r on the size of the net follows from finiteness of Vapnik Chervonenkis dimension. For general range spaces with finite VC dimension the logarithmic factor is necessary. However, there is by now several results that remove or reduce the logarithmic factor in "geometrically interesting range spaces" (as a sample, here is a result for axisparallel boxes). Unfortunately, as far as I know these developments do not include the space of triangles in the plane (I might be wrong; you should check with one of the experts).
Showing nonexistence of small 1/rnets for this problem is certainly open, and likely to be hard. The only currently known lower bound on 1/rnets for any geometric problem whatsoever is in my paper with Matoušek and Nivasch, and is for the family of all the convex sets. However, there instead of a grid {1,...,N}x{1,...,N} that is featured in this problem, the construction is a kind of highly stretched grid.

$\begingroup$ Could you clarify the relevance of the grid? I read through enough of you references to understand the statement that we are trying to build a "1/rnet of size O(r) for the family of all triangles with respect to the Lebesgue measure on [0,1]^2", but I don't understand why this is the same as building a "1/rnet for a grid {1,..,N}x{1,...,N} for large N". $\endgroup$ – David E Speyer Nov 1 '09 at 2:34

1$\begingroup$ If you can build an 1/rnet of size f(r) for the grid {1,...,N}x{1,...,N} for an arbitrarily large N for fixed r, then you can build an 1/rnet of size f(r) for the Lebesgue measure on [0,1]x[0,1]. That is because you can rescale the grid, and use compactness (the configuration space of positions of f(r) points in [0,1]^2 is compact). The point is that the number of point inside C in a sufficiently fine grid is almost proportional to the Lebesgue measure of C. $\endgroup$ – Boris Bukh Nov 1 '09 at 8:59
I've just spotted this question. It's a wellknown open problem in discrete geometry. Unfortunately, I can't remember what it is called.
There is a paper that proves that no union of finitely many lattices can work. I'll try to remember what it is. I remember that it used a result of Mahler.
Added later: I think it was Mahler's selection theorem. Maybe that's enough of a clue for someone to reconstruct the proof (which I seem to remember wasn't hugely hard).

$\begingroup$ I would be glad to learn of any places mentioning this problem. $\endgroup$ – Roland Bacher May 26 '10 at 18:23
This question is infuriating. I think I've made some progress, and would like to hear other's thoughts. Therefore, I am making this post a community wiki:
First of all, any triangle of area A contains a rectangle of area A/2. Proof: let the triangle be ABC, with AC the longest side. Let P and Q be the midpoints of AB and BC, and let R and S be the feet of perpendiculars from P and Q to AC. Then PQSR is a rectangle of the required area. Conversely, a rectangle of area A contains a triangle of area A/2. So we may instead ask whether there are large rectangles in the complement.
This is convenient because specifying a triangle involves 6 parameters, while specifying a rectangle has only 5. So this cuts our search space down a dimension. I find the most conveninent parameters for a rectangle to be the length of the longer side, L, the area, A, the angle of the longer side, $\theta$, and one of the vertices $(x,y)$. I'll call the type of the rectangle $(L, A, \theta)$, forgetting the translation parameters.
I now have a conjectural solution, although I have no idea how to prove that it works. I call this the sunflower configuration, because it was inspired by pattern of florets at the center of a sunflower.
Let $\tau$ be the Golden ratio. Consider the sequence of points $(\sqrt{k} \cos(2 \pi \tau k), \sqrt{k} \sin(2 \pi \tau k))$. A circle of radius $R$ around $0$ contains $R^2$ points, and has area $\pi R^2$, so the density is right.
Why do I think this is reasonable? There is a gorgeous property of the Golden ratio: if you look at the sequence $k \tau, (k+1) \tau, (k+2) \tau, ..., \ell \tau$ in $\mathbb R/\mathbb Z$, then the largest gap between any two consecutive angles is $\dfrac\tau{\ellk}$. So multiples of $2 \pi \tau$ are very well distributed around the unit circle. I learned this from Volume 2 of the Art of Computer Programming; I'll try to find a reference later if no one else does.
I don't have a strategy yet for proving that the sunflower pattern doesn't contain large triangles (or rectangles). But, where ever I try to place one, heuristics indicate that this equidistribution of angles destroys me.
My first strategy, aiming to prove a "no", was to overlay several lattices which were all rotations of each other. Let's fix $A$ once and for all, our goal is to exclude a rectangle of size $A$. I first set out to see when a lattice could contain a rectangle of type $(L, A, \theta)$.
Translate the rectangle so that one of the short sides touches $(0,0)$. Then the rectangle contains a circular wedge of radius $L$ and angle approximately $A/L^2$ with no lattice points. In other words, there is no $(p,q)$ with $\sqrt{p^2+q^2} \le L$ and $\theta  \tan^{1}(p/q) \le c\cdot A/L^2$, where $c$ is a constant I have not computed.
There are now a bunch of nuisances, having to do with the presence of that $tan^{1}$ and the fact that people who do Diophatine approximation usually ask for $q$ to be small, not $\sqrt{p^2+q^2}$. Passing over all of the details, the set of $\theta$'s for which such a rectangle exists should look something like the union of all intervals in the $L$th Farey sequence whose length is greater than $A/L^2$. Heuristics give me that the size of this is $c/A$ for $c$ a (different) constant that I haven't computed.
So, one lattice can't save me. The above suggests that, for every $L$, there will be a positive length set of $\theta$'s for which rectangles of type $(L, A, \theta)$ exist. Of course, we already knew that a single lattice couldn't work.
What about two lattices, rotated by some phi? I think I still lose. As $L$ grows, the set of theta's which work stays of size $1/A$, but becomes more and more spread out. Eventually, I would expect that it would contain two points that differ by phi. This part is very nonrigorous, though. In particular, I'm not sure whether we might be able to save things for some very special phi.
That's about how far I've gotten. I tried some other ideas, but didn't get anywhere. I'm not even sure which answer I think is right.
Concerning the nice construction David Speyer: I have the impression that it does not work. Reason: There are large regions where the configuration looks very much like a lattice with fairly large density (no close points). Such a region contains thus large triangles (one can of course replace triangles by convex set or by rectangles, this does not change the problem). I did a fairly coarse and rapid computation which seems to show that there are indeed regions contained in arbitrarily large discs where the points of the set are at most at distance \epsilon (for arbitrary small positive \epsilon) from the points of a dense lattice. This computation (if correct) implies easily the existence of arbitrarily large triangles in the complement.
Concerning the history of the problem: I am not aware that it has been considered in the present form. I have written a paper "Universal Convex Coverings" which is appearing in the Bulletin of the LMS (a copy is on the arXiv) where I construct a set with R^2 log(R) points of (Euclideean) norm at most R which intersects all triangles (or equivalently, convex sets) of large enough area. (The paper contains in fact a construction for arbitrary dimensions). I am thus curious if one can get rid of the logarithm. In fact, I believe that the logarithm can be improved to some smaller slowly increasing function but I would be very surprised by the existence of such a set with uniform density.

$\begingroup$ I don't understand, Roland. Why do you say "There are large regions where the configuration looks very much like a lattice with fairly large density" about David's proposed example? $\endgroup$ – Scott Morrison♦ Nov 1 '09 at 0:33

$\begingroup$ I would also like more detail. On which region, and for what lattice, is this true? $\endgroup$ – David E Speyer Nov 1 '09 at 1:49
This seems somewhat related to Heilbronn's triangle problem, which is about determining the worstcase behavior of the minimum area triangle determined by a set of n points in a unit disk. In this one, instead of the smallest triangle determined by a set of points, we want the largest empty triangle; I think an equivalent version of the question is to ask whether or not, for all sets of n points in the unit disk, the largest empty triangle has area O(1/n).
However unlike Heilbronn's problem the triangle we want isn't required to have three of the n given points as its vertices; for instance, if n points were placed close together within the disk there would be no large triangle having three of them as vertices but there would exist other large empty triangles. I did at least find one paper that's sort of related to this: On a Triangle with the Maximum Area in a Planar Point Set, Hisono et al 2005, show that there exist sets of n points for which the ratio of areas of the largest empty triangle with three of the points as vertices and the convex hull of all the vertices is O(1/n).
By the way, it's easy to define sets of points such that there is no large empty triangle containing the origin: place a ring of 2^k points at radius 2^k, for k = 1, 2, 3, 4, ... The density of points within any nonconstantarea disk will then be upper and lower bounded by a constant times the area of the disk, and by adding other points one can make the upper and lower bound constants as close as one wishes to each other. So somehow, if the large triangles always exist, they must sometimes be far from the origin.
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 QuasiMonte 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 Wikipediadiscrepancy.) The results are typically expressed in terms of "discrepancy": for N points in the unit square, the discrepancy D_{N} can be defined as
sup_{J} 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 D_{N} as a function of N. For the axisoriented rectangles, it is Ω(log N), and this can be achieved by the HaltonHammersley sequence. When J is changed to rectangles with any rotation, discrepancy is Ω(N^{1/4}), and O(N^{1/4} √log N) can be achieved by jittered sampling. When J is all convex subsets of the unit square, you've moved to "isotropic discrepancy" J_{N}  harder to study, but D_{N} ≤ J_{N} ≤ 8√(N D_{N}). Also, J_{N} is Ω(N^{1/4}).
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 j_{1} with n_{1} points, and add an empty region (while staying in the set J), you can show that the empty region must have area O(2D_{N}), 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.

$\begingroup$ Thanks for this information which seems to be an analogue of epsilonnets. I have the impression that these problems are slightly different due to the fact that one considers compact sets. There is however an intriguing "log" appearing in a conjecture which I encountered also in my examples. I need probably some time for thinking (which I lack unfortunately at the moment) for understanding this better. $\endgroup$ – Roland Bacher Aug 27 '10 at 13:45
This answer concerns the search for grids in the sunflower model. My confidence is shaken but not broken.
I think I have figured out what Roland is saying. Consider a square of area A, at distance R from the origin, with R very large. It will be convenient to identify our plane with the complex plane. So our points are of the form e^{(1/2) log k + 2 pi i tau k}.
Let L(z) = R log z. On our little square, L is approximately area preserving and approximately linear, with error O(1/R). So our square will contain large lattice point free triangles if and only if L of our square does.
Well, L of the sunflower pattern is the points of the form (1/2) R log k + 2 pi i tau R k. We only care about k between (R\sqrt{A})^2 and (R + \sqrt{A})^2 (roughly), which is to say k of the form R^2 +s for s < 2 \sqrt{A} R. For such k,
(1/2) R log k + 2 pi i tau R k = R log R + (1/2) s/R + 2 pi i tau R^3 + 2 pi i tau s R = c + (1/2) s/R + 2 pi i tau s R,
where c does not depend on s. Of course, log is multivalued, so the right answer is actually
c + s (1/2R + 2 pi i tau R) + t (2 pi i R).
In short, the sunflower pattern, restricted to our square, looks like the intersection of a similar square with a translate of the grid generated by (1/2R + 2 pi i tau R) and (2 pi i R). I notice this has fundamental domain 2 pi, so I am guessing this is the grid that Roland is discussing.
So, why am I not convinced this is a problem? The shape of the grid is changing with R. Basically, we are intersecting our square (of area A) with a grid whose fundamental regions are getting very thin and tall. It is not clear to me how we can find large triangles INSIDE the square of area A.
In fact, that would be a great question for some one to answer. Take a square of area 100. Intersect it with the grid spanned by (1/R, tau R) and (0, R). Let f(R) be the largest triangle inside the square, missing this lattice. As R goes to infty, does f(R) approach 1, or stay nearer to 100? (I have removed factors of 2 and pi that I expect to be nonessential.)
Of course, I may have misinterpreted Roland. Roland, which grid is close to the sunflower model?
Here are a few details for proving that the construction of David Speyer does not work:
Consider an arbitrarily large "window", say $$W=\lbrace(x,y)A100\leq x\leq A+100,B100\leq x \leq B+100]\rbrace$$ (one has to replace 100 by arbitrarily large numbers later) centered at some point $(A,B)$ of the plane. If $(A,B)$ is far from the origin, the intersection of $W$ with the set $S$ of all points of the form $(\sqrt{k}\cos(2k\pi),\sqrt{k}\sin(2k\pi)), k\in\mathbb N$, looks (generically, but one can neglect the nongeneric case since it fits into the generic picture also) "almost" like a translate $\Lambda$ of a lattice with a fundamental domain of area $\pi$. This translate $\Lambda$ of a lattice depends of course on the precise location of $(A,B)$ but the "convergency" of $W\cap S$ to such a $\Lambda$ depends only on the norm of (A,B) and on the size of W. More precisely, convergency means that for any strictly positive epsilon there exists a real number $R=R(epsilon,100)$ depending only on epsilon and the size (here 100) of the window with the property that if $(A,B)$ has norm at least $R$, then there exists a translate $\Lambda$ (depending on $(A,B)$) of an Euclidean lattice with fundamental domain $\pi$ and a map from $S\cap W$ (the points of $S$ in the window $W$) into $\Lambda$ which moves points at most by $\epsilon$. For $\epsilon$ small enough (depending on nothing) there are arbitrarily large triangles not intersecting an $\epsilon$neighbourhood of such a set $\Lambda$. For $\epsilon$ small enough, the diameter of such triangles of area $\alpha$ can be bounded above by some real number depending only on the area \alpha of the triangle but not on the translate $\Lambda$ of an Euclidean lattice with fundamental domain $\pi$. Given a real number $\alpha$, we can thus choose the window $W$ large enough such that it contains triangles of area $\alpha$ which avoid an $\epsilon$neighbourhood of $\Lambda$. Such a triangle avoids thus $S$.
I hope that these explanations are sufficiently explicit. All this is essentially only elementary lattice theory and nothing is difficult to prove.
I don't know that I'm quite as hooked on this problem as David S., but it really is addictive. My hunch is also that the answer should be "no," and I really really want this to be solvable by simple (if clever) combinatorialgeometry arguments. Here's some arguments of that nature which I don't think can be themselves made to go through, but which do rule out some approaches. I'm making the post community wiki, so people can add to it or correct the stupid mistakes which I've almost certainly made.
So we'll use the rectangle formulation. Suppose that S is a set of points that does intersect every sufficiently large triangle. Then if we project S onto any line, the image has to be dense. This is doable; I think that David's tauconstruction has this property. But you can't build such a set up out of lattices in an easy way.
But we can do slightly better than that, although it's harder to state. You also need the projections of S intersect a strip parallel to a line L onto L to be very nicely distributed. In particular, the maximum distance between two consecutive points has to be O(1/x) for a strip of width x, which is best possible up to a constant. (To see this, notice that there are O(Lx) points in the strip, and they have to cover an interval of length L.).
By the way, this subsumes the appeals to "embedding large almostlattices." I'd (harrison) be interested to see a set that has this stronger projection property but provably contains large rectangles. ETA: Upon further reflection, if the implied constant is uniform for all lines L, then this property's necessary and sufficient, since the projection property bounds the size of a rectangle in the complement of S that's parallel to L.
It is important to note that this rules out most straightforward random constructions. If we place N points at random on a line segment of length 1, we anticipate a gap of length log N/N somewhere. We need to do better, achieving gaps of size O(1/N). Proof: Divide the segment into N buckets. For any k consecutive buckets, the chance that they are all missed is (1k/n)^n ~ e^{k}. So, if k is log N + O(1), the chance that a particular k buckets are missed will be of order 1/N. Since there are on the order of N blocks of k consecutive buckets, the expected number of such blocks which are missed will be positive. So there is a positive probability of missing a segment of length log N/N.
David, considering the logarithm is a nice idea. A small error: the fundamental domain has area pi: There is a factor of 2 in the denominator of the real part.
You are of course right: The area of all triangles (not intersecting sunflower points) contained in squares (or windows) of fixed area A is bounded. One has however to consider the limit when A goes to infinity. Then your argument is no longer valid and you get indeed arbitrarily large triangles not intersecting the sunflower points in sufficiently large squares sufficiently far away from the origin.
Your interpretation is correct: (Up to the logarithm,) this is indeed the grid (I prefer to speak of a translate of an Euclidean lattice).
Conclusion: A solution to the problem (if it exists) should have no arbitrarily large regions where it looks very much like a Euclidean lattice (which is the case for the sunflower point set).
Another nice idea which unfortunately fails to work is to take the set S of all vertices (or barycenters or ???) of all triangles involved in a Penrose tiling (giving those beautiful quasiperiodic pictures with almost symmetries of order five). Indeed, such tilings contain arbitrarily long straight lines. This implies the existence of triangles with arbitrarily large area which do not intersect S.
I mentioned a necessary and sufficient condition that a set must satisfy to intersect all large triangles. Here I'll describe a set that satisfies a (very) weak version of that condition  namely, the projections of strips around the xaxis satisfy the appropriate property. Actually I'll describe the construction on [0, 1) x [0, \infty); we can translate to get the whole xaxis. I'm not optimistic that this can be turned into something which solves the problem, but I'll describe it anyway since it's hopefully illustrative.
The key thing is that binary fractions m/2^k are dense in the reals, and the ones between 0 and 1 have a simple ordering which allows us to satisfy the density condition. The construction is as follows: Place a point at (0, 0), and for every positive integer m with binary expansion 1a_2...a_k, place a point at (0.a_k...a_2 1, m). So for instance there are points at (1/2, 1), and at (3/8, 6). Then a square of side length n has at most n^2 points within it, for obvious reasons. The projection property of strips R x [0, y) onto the xaxis is easily checked as well. (Question: Does this satisfy the projection property for strips parallel to the xaxis? I think it might, essentially because this is related to 2adicness.)
Obviously we can do something similar for any prime p replacing 2. I'd be interested to see, though, if someone can provide an appropriate construction of this type for the "infinite prime"  does it have similarly nice properties?
I think I can rule out some really long rectangles, in David's sunflower model.
In particular, if one corner of the rectangle is with \sqrt{R}
of the origin, then the rectangle has side length at most R
. I'm not sure this is at all useful, however; polymath style, someone can point that out for me!
The argument is just the following. Let's write Θ(x,y,r)
for the angular width of the widest empty wedge with base at (x,y)
and radius r
. The important property of the sunflower model is that the widest wedge at the origin with radius r
has width at most c/r^2
, for some constant c
, i.e. Θ(0,0,r)
≤ c/r^2
. Now let's move to an arbitrary point (x,y)=(r
_{0}cos
θ, r
_{0}sin
θ)
. The angle to a point P
differs from the angle from the origin to P
by at most r
_{0}/2R
, where R
is the distance from the origin to P
. Thus
Θ(r
_{0}cos
θ, r
_{0}sin
θ, R')
≤ Θ(0,0,R'r
_{0}) + r
_{0}/(R'r
_{0})
.
As long as R'
≥r
_{0}^2
, this is small enough that we don't need to worry. As it's the quantity R
Θ(x,y,R)
that we need to bound, unfortunately I think this is the best you can do without using anything stronger about the distribution of points.

$\begingroup$ Actually, this didn't use the sunflower model at all  as long as you can't have arbitrarily large rectangles at the origin, this estimate works. That makes it sound less interesting, I guess. $\endgroup$ – Scott Morrison♦ Nov 1 '09 at 0:30