The number of unit-area triangles in the plane: Theme and variations1footnote 11footnote 1Work on this paper by Orit E. Raz and Micha Sharir was supported by Grant 892/13 from the Israel Science Foundation and by the Israeli Centers of Research Excellence (I-CORE) program (Center No. 4/11). Work by Micha Sharir was also supported by Grant 2012/229 from the U.S.–Israel Binational Science Foundation and by the Hermann Minkowski-MINERVA Center for Geometry at Tel Aviv University. Part of this research was performed while the authors were visiting the Institute for Pure and Applied Mathematics (IPAM), which is supported by the National Science Foundation.

The number of unit-area triangles in the plane: Theme and variations111Work on this paper by Orit E. Raz and Micha Sharir was supported by Grant 892/13 from the Israel Science Foundation and by the Israeli Centers of Research Excellence (I-CORE) program (Center No. 4/11). Work by Micha Sharir was also supported by Grant 2012/229 from the U.S.–Israel Binational Science Foundation and by the Hermann Minkowski-MINERVA Center for Geometry at Tel Aviv University. Part of this research was performed while the authors were visiting the Institute for Pure and Applied Mathematics (IPAM), which is supported by the National Science Foundation.

Orit E. Raz School of Computer Science, Tel Aviv University, Tel Aviv 69978, Israel. oritraz@post.tau.ac.il    Micha Sharir School of Computer Science, Tel Aviv University, Tel Aviv 69978, Israel. michas@post.tau.ac.il
Abstract

We show that the number of unit-area triangles determined by a set of points in the plane is , improving the earlier bound of Apfelbaum and Sharir [2]. We also consider two special cases of this problem: (i) We show, using a somewhat subtle construction, that if consists of points on three lines, the number of unit-area triangles that spans can be , for any triple of lines (it is always in this case). (ii) We show that if is a convex grid of the form , where , are convex sets of real numbers each (i.e., the sequences of differences of consecutive elements of and of are both strictly increasing), then determines unit-area triangles.

1 Introduction

In 1967, Oppenheim (see [9]) asked the following question: Given points in the plane and , how many triangles spanned by the points can have area ? By applying a scaling transformation, one may assume and count the triangles of unit area. Erdős and Purdy [8] showed that a section of the integer lattice determines triangles of the same area. They also showed that the maximum number of such triangles is at most . In 1992, Pach and Sharir [10] improved the bound to , using the Szemerédi-Trotter theorem [16] (see below) on the number of point-line incidences. More recently, Dumitrescu et al. [4] have further improved the upper bound to , by estimating the number of incidences between the given points and a 4-parameter family of quadratic curves. In a subsequent improvement, Apfelbaum and Sharir [2] have obtained the upper bound , for any , which has been slightly improved to in Apfelbaum [1]. This has been the best known upper bound so far.

In this paper we further improve the bound to . Our proof uses a different reduction of the problem to an incidence problem, this time to incidences between points and two-dimensional algebraic surfaces in . A very recent result of Solymosi and De Zeeuw [15] provides a sharp upper bound for the number of such incidences, similar to the Szemerédi–Trotter bound, provided that the points, surfaces, and incidences satisfy certain fairly restrictive assumptions. The main novel features of our analysis are thus (a) the reduction of the problem to this specific type of incidence counting, and (b) showing that the assumptions of [15] are satisfied in our context.

After establishing this main result, we consider two variations, in which better bounds can be obtained.

We first consider the case where the input points lie on three arbitrary lines. It is easily checked that in this case there are at most unit-area triangles. We show, in Section 3, that this bound is tight, and can be attained for any triple of lines. Rather than just presenting the construction, we spend some time showing its connection to a more general problem studied by Elekes and Rónyai [6] (see also the recent developments in [7, 11, 12]), involving the zero set of a trivariate polynomial within a triple Cartesian product. Skipping over the details, which are spelled out in Section 3, it turns out that the case of unit-area triangles determined by points lying on three lines is an exceptional case in the theory of Elekes and Rónyai [6], which then leads to a construction with unit-area triangles.

Another variation that we consider concerns unit-area triangles spanned by points in a convex grid. That is, the input set is of the form , where and are convex sets of real numbers each; a set of real numbers is called convex if the differences between consecutive elements form a strictly increasing sequence. We show that in this case determine unit-area triangles. The main technical tool used in our analysis is a result of Schoen and Shkredov [13] on difference sets involving convex sets.222Very recently, in work in progress, jointly with I. Shkredov, the bound is further improved in this case.

2 Unit-area triangles in the plane

Theorem 1.

The number of unit-area triangles spanned by points in the plane is .

We first recall the Szemerédi–Trotter theorem [16] on point-line incidences in the plane.

Theorem 2 (Szemerédi and Trotter [16]).

(i) The number of incidences between distinct points and distinct lines in the plane is . (ii) Given distinct points in the plane and a parameter , the number of lines incident to at least of the points is . Both bounds are tight in the worst case.

Proof of Theorem 1. Let be a set of points in the plane, and let denote the set of unit-area triangles spanned by . For any pair of distinct points, , let denote the line through and . The points for which the triangle has unit area lie on two lines parallel to and at distance from on either side. We let be the line that lies to the left of the vector . We then have

 |U|=13∑(p,q)∈S×S,p≠q|ℓ′pq∩S|.

It suffices to consider only triangles of , that have the property that at least one of the three lines , , is incident to at most points of , because the number of triangles in that do not have this property is . Indeed, by Theorem 2(ii), there exist at most lines in , such that each contains at least points of . Since every triple of those lines supports (the edges of) at most one triangle (some of the lines might be mutually parallel, and some triples might intersect at points that do not belong to ), these lines support in total at most triangles, and, in particular, at most triangles of . Since this number is subsumed in the asserted bound on , we can therefore ignore such triangles in our analysis. In what follows, denotes the set of the remaining unit-area triangles.

We charge each of the surviving unit-area triangles to one of its sides, say , such that contains at most points of . That is, we have

 |U|≤∑(p,q)∈(S×S)∗|ℓ′pq∩S|,

where denotes the subset of pairs , such that , and the line is incident to at most points of .

A major problem in estimating is that the lines , for , are not necessarily distinct, and the analysis has to take into account the (possibly large) multiplicity of these lines. (If the lines were distinct then would be bounded by the number of incidences between lines and points, which is — see Theorem 2(i).) Let denote the collection of lines (without multiplicity). For , we define to be the set of all pairs , with , and for which . We then have

 |U|≤∑ℓ∈L|ℓ∩S||(S×S)ℓ|.

Fix some integer parameter , to be set later, and partition into the sets

 L−={ℓ∈L∣|ℓ∩S|n/k}.

We have

 |U|≤∑ℓ∈L−|ℓ∩S||(S×S)ℓ|+∑ℓ∈L+|ℓ∩S||(S×S)ℓ|+∑ℓ∈L++|ℓ∩S||(S×S)ℓ|.

The first sum is at most , because is at most . The same (asymptotic) bound also holds for the the third sum. Indeed, since , the number of lines in is at most , as follows from Theorem 2(ii), and, for each , we have and (for any , , there exists at most one point , such that ). This yields a total of at most unit-area triangles. It therefore remains to bound the second sum, over .

Applying the Cauchy-Schwarz inequality to the second sum, it follows that

Let (resp., ), for , denote the number of lines for which (resp., ). By Theorem 2(ii), . Hence

 ∑ℓ∈L+|ℓ∩S|2 =n/k∑j=kj2Nj≤k2N≥k+n/k∑j=k+1(2j−1)N≥j =O⎛⎝n2k+nk+n/k∑j=k+1(n2j2+n)⎞⎠=O(n2k)

(where we used the fact that ). It follows that

 |U|=O(n2k+nk1/2(∑ℓ∈L+|(S×S)ℓ|2)1/2).

To estimate the remaining sum, put

 Q:={(p,u,q,v)∈S4∣(p,u),(q,v)∈(S×S)ℓ, for % some ℓ∈L+}.

That is, consists of all quadruples such that , and each of contains at most points of . See Figure 1(a) for an illustration.

The above bound on can then be written as

 |U|=O(n2k+n|Q|1/2k1/2). (1)

The main step of the analysis is to establish the following upper bound on .

Proposition 3.

Let be as above. Then

The proposition, combined with (1), implies that , which, if we choose , becomes . Since the number of triangles that we have discarded is only , Theorem 1 follows. ∎

Proof of Proposition 3. Consider first quadruples , with all four points collinear. As is easily checked, in this case must also satisfy . It follows that a line in the plane, which is incident to at most points of , can support at most such quadruples. By definition, for each , so the line is incident to at most points of , and it suffices to consider only lines with this property. Using the preceding notations , , the number of quadruples under consideration is

 O⎛⎜⎝∑j≤n1/2j3Nj⎞⎟⎠=O⎛⎜⎝∑j≤n1/2j2N≥j⎞⎟⎠=O⎛⎜⎝∑j≤n1/2j2⋅n2j3⎞⎟⎠=O(n2logn).

This is subsumed by the asserted bound on , so, in what follows we only consider quadruples , such that are not collinear.

For convenience, we assume that no pair of points of share the same - or -coordinate; this can always be enforced by a suitable rotation of the coordinate frame. The property that two pairs of are associated with a common line of can then be expressed in the following algebraic manner.

Lemma 4.

Let , and represent , and , by their coordinates in . Then if and only if

 y−bx−a=w−dz−candbx−ay+2x−a=dz−cw+2z−c. (2)

Proof. Let be such that . Then, by the definition of , we have

 12∣∣ ∣∣axtbyαt+β111∣∣ ∣∣=1,

or

 (b−y−α(a−x))t−β(a−x)+ay−bx=2,

for all . Thus,

 α =α(a,b,x,y)=y−bx−a, β =β(a,b,x,y)=bx−ay+2x−a.

Then the constraint can be written as

 α(a,b,x,y) =α(c,d,z,w), β(a,b,x,y) =β(c,d,z,w),

which is (2). ∎

We next transform the problem of estimating into an incidence problem. With each pair , we associate the two-dimensional surface which is the locus of all points that satisfy the system (2). The degree of is at most , being the intersection of two quadratic hypersurfaces. We let denote the set of surfaces

 Σ:={σpq∣(p,q)∈(S×S)∗}.

For , the corresponding surfaces are distinct. The proof of this fact is not difficult, but is somewhat cumbersome, and we therefore omit it, since our analysis does not use this property. We also consider the set regarded as a point set in (identifying ). We have . The set , the set of incidences between and , is naturally defined as

 I(Π,Σ):={(π,σ)∈Π×Σ∣π∈σ}.

By Lemma 4, we have , where and . This implies that .

Consider the subcollection of incidences , such that are non-collinear (as points in ). As already argued, the number of collinear quadruples in is , and hence . So to bound it suffices to obtain an upper bound on .

For this we use the following recent result of Solymosi and De Zeeuw [15] (see also the related results in [14, 17]). To state it we need the following definition, which is a specialized version of the more general original definition in [15].

Definition 5.

A two-dimensional constant-degree surface in is said to be slanted (the original term used in [15] is good), if, for every , is finite, for , where and are the projections of onto its first and last pairs of coordinates, respectively.

Theorem 6 (Solymosi and De Zeeuw [15]).

Let be a subset of , and let be a finite set of two-dimensional constant-degree slanted surfaces. Set , and let . Assume that for every pair of distinct points there are at most surfaces such that both pairs are in . Then

 |I|=O(|Π|2/3|Σ|2/3+|Π|+|Σ|).

To apply Theorem 6, we need the following key technical proposition, whose proof is given in the next subsection.

Proposition 7.

Let , , and be the sets that arise in our setting, as specified above. Then, (a) the surfaces of are all slanted, and (b) for every pair of distinct points , there are at most three surfaces such that both pairs are in .

We have . Therefore, Theorem 6 implies that , which completes the proof of Proposition 3 (and, consequently, of Theorem 1). ∎

2.1 Proof of Proposition 7

We start by eliminating and from (2). An easy calculation shows that

 z =2(x−a)(b−d)(x−a)+(c−a)(y−b)+2+c, (3) w =2(y−b)(b−d)(x−a)+(c−a)(y−b)+2+d.

This expresses as the graph of a linear rational function from to (which is undefined on the line at which the denominator vanishes). Passing to homogeneous coordinates, replacing by and by , we can re-interpret as the graph of a projective transformation , given by

The representation (3) implies that every defines at most one pair such that . By the symmetry of the definition of , every pair also determines at most one pair such that . This shows that, for any , the surface is slanted, which proves Proposition 7(a).

For Proposition 7(b), it is equivalent, by the symmetry of the setup, to prove the following dual statement: For any , such that , we have .

Let be as above, and assume that . Note that this means that the two projective transformations , agree in at least four distinct points of the projective plane. We claim that in this case and , regarded as graphs of functions on the affine -plane, must coincide on some line in that plane.

This is certainly the case if and coincide. (As mentioned earlier, this situation cannot arise, but we include it since we did not provide a proof of its impossibility.) We may thus assume that these surfaces are distinct, which implies that and are distinct projective transformations.

As is well known, two distinct projective transformations of the plane cannot agree at four distinct points so that no three of them are collinear. Hence, out of the four points at which and agree, three must be collinear. Denote this triple of points (in the projective -plane) as , and their respective images (in the projective -plane) as , for . Then the line that contains , , is mapped by both and to a line , and both transformations coincide on (since they both map the three distinct points , , to the same three respective points , , ).

Passing back to the affine setting, let then be a pair of lines in the -plane and the -plane, respectively, such that, for every (other than the point at which the denominator in (3) vanishes) there exists , satisfying . We show that in this case are all collinear and .

We first observe that . Indeed, if each of and is either empty or infinite, then we must have (since both are parallel to ). Otherwise, assume without loss of generality that , and let denote the unique point in this intersection. Let be the point such that satisfies (3) with respect to both surfaces , (the same point arises for both surfaces because ). That is, , and . In particular, , and . Since, by construction, , we have , which yields that also . Thus necessarily are collinear, and , as claimed.

Assume that at least one of , intersects in exactly one point; say, without loss of generality, it is , and let denote the unique point in this intersection. Similar to the argument just made, let be the point such that satisfies (3) with respect to both surfaces , . Note that since , we must have too, and . In particular, since , by assumption, we also have . Using the properties and , it follows that the triangles , are congruent; see Figure 2(a). Thus, in particular, . Since, by construction, also , it follows that . We conclude that in this case are collinear and .

We are therefore left only with the case where each of and is either empty or infinite. That is, we have (since both are parallel to ). As has already been argued, we also have , and thus is a parallelogram; see Figure 2(b). In particular, . Let be the intersection point of with , and let be the point such that satisfies (3) with respect to both surfaces , . By construction and . Hence must lie on . It is now easily checked that the only way in which can lie on both surfaces and is when are all collinear; see Figure 2(b).

To recap, so far we have shown that for , , , and as above, either , or , , , and are collinear with . It can then be shown that, in the latter case, any point must satisfy ; see Figure 1(b). Thus, for a point incident to each of , , neither of , is in . In other words, in this case. This contradiction completes the proof of Proposition 7.

3 Unit-area triangles spanned by points on three lines

In this section we consider the special case where is contained in the union of three distinct lines , , . More precisely, we write , with , for , and we are only interested in the number of unit-area triangles spanned by triples of points in . It is easy to see that in this case the number of unit-area triangles of this kind is . Indeed, for any pair of points , the line intersects in at most one point, unless coincides with . Ignoring situation of the latter kind, we get a total of unit-area triangles. If no two lines among are parallel to one another, it can be checked that the number of pairs such that is at most a constant, thus contributing a total of at most unit-area triangles. For the case where two (or more) lines among are parallel, the number of unit-area triangles is easily seen to be .

In this section we present a rather subtle construction that shows that this bound is tight in the worst case, for any triple of distinct lines. Instead of just presenting the construction, we spend some time showing its connection to a more general setup considered by Elekes and Rónyai [6] (and also, in more generality, by Elekes and Szabó [7]).

Specifically, the main result of this section is the following.

Theorem 8.

For any triple of distinct lines in , and for any integer , there exist subsets , , , each of cardinality , such that spans unit-area triangles.

Proof. The upper bound has already been established (for any choice of ), so we focus on the lower bound. We recall that by the area formula for triangles in the plane, if

 12∣∣ ∣ ∣∣pxqxrxpyqyry111∣∣ ∣ ∣∣=1, (4)

then the points , and form the vertices of a positively oriented unit-area triangle in . (Conversely, if has area 1 then the left-hand side of (4) has value , depending on the orientation of ..)

To establish the lower bound, we distinguish between three cases, depending on the number of pairs of parallel lines among .

The three lines l1,l2,l3 are mutually parallel.

In this case we may assume without loss of generality that they are of the form

 l1 ={(t,0)∣t∈R}, l2 ={(t,1)∣t∈R}, l3 ={(t,α)∣t∈R},

for some . (We translate and rotate the coordinate frame so as to place at the -axis and then apply an area-preserving linear transformation that scales the - and -axes by reciprocal values.) We set

 S1 :={(xi:=i1−α,0)∣i=1,…,n}⊂l1, S2 :={(yj:=jα,1)∣j=1,…,n}⊂l2, S3 :={(zij:=i+j−2,α)∣i,j=1,…,n}⊂l3.

Clearly each of the sets , , is of cardinality . Note that for every pair of indices , we have By (4), every such pair corresponds to a unit-area triangle with vertices , and . That is, spans unit-area triangles.

There is exactly one pair of parallel lines among l1,l2,l3.

Using an area-preserving affine transformation333In more generality than the transformation used in the first case, these are linear transformations with determinant . of (and possibly re-indexing the lines), we may assume that

 l1 ={(t,0)∣t∈R}, l2 ={(t,1)∣t∈R}, l3 ={(0,t)∣t∈R}.

We claim that in this case the sets

 S1 :={(xi:=2i+2,0)∣i=1,…,n}⊂l1, S2 :={(yj:=2j+2,1)∣j=1,…,n}⊂l2, S3 :={(0,zij:=11−2j−i)∣i,j=1,…,n,i≠j}⊂l3,

span unit-area triangles. As before, , and are each of cardinality .

Using (4), the triangle spanned by , , and has unit area if

 12∣∣ ∣ ∣∣xiyj001zij111∣∣ ∣ ∣∣=1,

or

 xi−zij(xi−yj)2=1,

or

 zij=xi−2xi−yj=11−yj−2xi−2.

Since the latter holds for every , we get unit-area triangles, as claimed.

No pair of lines among l1,l2,l3 are parallel.

This is the most involved case. Using an area-preserving affine transformation of (that is, a linear map with determinant and a translation), we may assume that the lines are given by

 l1 ={(t,0)∣t∈R}, l2 ={(0,t)∣t∈R}, l3 ={(t,−t+α)∣t∈R},

for some . By (4) once again, the points , , and span a unit-area triangle if