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Extremal problems for pairs of triangles in a convex polygon
Zoltan Furedi* Dhruv Mubayi Jason O’Neill Jacques Verstraete
October 21, 2020
Abstract
A convex geometric hypergraph or cgh consists of a family of subsets of a strictly convex set of
points in the plane. The study of cghs is motivated by problems in combinatorial geometry, and was
studied at length by Braß and by Aronov, Dujmovic, Morin, Ooms and da Silveira. In this paper,
we determine the extremal functions for five of the eight configurations of two triangles exactly and
another one asymptotically. We give conjectures for two of the three remaining configurations. Our
main results solve problems posed by Frankl, Holmsen and Kupavskii on intersecting triangles in
a cgh. In particular, we determine the exact maximum size of an intersecting family of triangles
whose vertices come from a set of n points in the plane.
1 Introduction
A convex geometric hypergraph or cgh is a family of subsets of a set of points in strictly convex position
in the plane – we assume these points, denoted by Ωn = v0, v1, . . . , vn−1, are the vertices of some
regular n-gon with the clockwise cylic ordering v0 < v1 < · · · < vn−1 < v0. For an r-uniform cgh F –
an r-cgh for short – let the extremal function ex(n, F ) denote the maximum number of edges in an
r-uniform cgh on n points that does not contain F – an F -free cgh. For the rich history of ordered
and convex geometric graph problems and their applications, see Tardos [27] and Pach [21, 22] and
for cghs, see Braß [3].
In this paper, we concentrated on intersection patterns of pairs of triangles. For r = 3, there are eight
configurations of two triangles (we refer interchangeably to triangles and triples or edges when r = 3),
depicted below:
*Alfred Renyi Institute of Mathematics, Hungarian Academy of Sciences, P.O. Box 127, Budapest, Hungary, H-1364.
Research was supported in part by NKFIH grant KH130371 and NKFI–133819. E-mail: [email protected] of Mathematics, Statistics, and Computer Science, University of Illinois at Chicago. Research supported
by NSF award DMS-1952767. E-mail: [email protected] of Mathematics, University of California, San Diego. Research supported by NSF award DMS-1800332.
E-mail: [email protected] and [email protected]
1
Figure 1
The configurations M1,M2 and M3 when viewed as hypergraphs are matchings with two edges, whereas
S1, S2 and S3 are stars. For all of these configurations, Braß [3] has shown the extremal function is
either Θ(n2) or Θ(n3); the latter arises precisely when the two triangles have no common interior
point. Aronov, Dujmovic, Morin, Ooms and da Silveira [1] extensively studied cghs which avoid
combinations of these pictures, and determined many of the order of magnitudes of the associate
extremal number. Furthermore, certain combinations of the pictures above are closely related to
problems in combinatorial geometry, extremal matrix theory, and combinatorial number theory [1]. In
particular, the extremal problem for forbidding both S2 and D2 is tightly connected to the monotone
matrix problem, to the tripod packing problem and 2-comparable triples problem. Lower bounds on the
extremal function in these cases were given by Gowers and Long [12], and upper bounds come from the
triangle removal lemma, however these bounds remain far apart. Extremal problems for matchings in
ordered graphs connect to enumeration of permutations [19] and extensions to hypergraphs [15]. In
recent work, Frankl, Holmsen and Kupavskii [8] determined the maximum number of triangles formed
by an n-point convex set that pairwise intersect geometrically. In other words, referring to Figure
1, they studied ex(n, M1, S1, D1). In this paper, for every configuration F in Figure 1 we prove
bounds or exact results for ex(n, F ). Most of the theorems hold equally for ordered hypergraphs,
where the vertex set is linearly ordered.
1.1 Configurations M1, S1 and D1. Frankl, Holmsen and Kupavskii [8] raised the problem of
determining the maximum size of an n-vertex cgh not containing two triangles with disjoint interiors.
This corresponds to forbidding all three configurations M1, S1 and D1. The principal extremal con-
struction for M1, S1 and D1 of Frankl, Holmsen and Kupavskii [8] is as follows. All constructions have
vertex set Ωn = v0, v1, . . . , vn−1 with the convention that subscripts are always mod n and we write
2
vi < vj < vk to denote that in the clockwise cyclic ordering of Ωn, vi precedes vj which precedes vk.
Construction 1. For n ≥ 3 odd, let H?(n) comprise the single cgh consisting of triangles which
contain in their interior the centroid of Ωn. For n ≥ 4 even, each H ∈ H?(n) consists of all triangles
which contain the centroid of Ωn and, for each diameter vi, vi+n/2 of Ωn, we either add all triangles
vi, vj , vi+n/2 where vi < vj < vi+n/2, or all triangles vi, vj , vi+n/2 where vi+n/2 < vj < vi.
Let ·4(n) be the common size of each H ∈ H?(n). Then it is straightforward to show (see [8]) that
·4(n) =
n(n− 1)(n + 1)
24if n is odd
n(n− 2)(n + 2)
24if n is even.
In particular, ·4(n)/(n3
)→ 1/4 as n→∞. It is also straightforward to check that no two triangles in
any H ∈ H?(n) have disjoint interiors, so
ex(n, M1, S1, D1) ≥ ·4(n).
Frankl, Holmsen and Kupavskii [8] proved that equality holds above and their proof moreover shows
that any extremal M1, S1, D1-free cgh is one of those described in Construction 1. They posed the
following problem (see Problem 2 in [8]).
Problem 1. What happens if one relaxes the intersecting condition and allows triangles to intersect
on the boundary?
If we allow two triangles to intersect on any part of the boundary, then we allow S1 and D1 so Problem
1 asks for ex(n,M1). If, on the other hand, we allow boundary intersections but only on an edge
boundary, then we allow only D1, so Problem 1 asks for ex(n, M1, S1). We will determine both
ex(n,M1) and ex(n, M1, S1) exactly. First we describe two more constructions.
Construction 2. For n ≥ 3 odd, the unique cgh in H′(n) is obtained by adding all triangles containing
a pair vi, vi+(n−1)/2 to any cgh in H?(n) (left diagram in Figure 2). For n ≥ 4 even, H′(n) is obtained
by adding all triangles containing a diameter of Ωn, plus all triangles containing a pair from a set of
n/2 pairwise intersecting pairs of the form vi, vi+n/2−1 (right diagram in Figure 2).
Figure 2: Construction of H′(n)
3
By inspection, every cgh in H′(n) is M1-free, and has size ·4(n) + n(n− 3)/2.
Construction 3. For n ≥ 3 odd, each cgh in H+(n) is obtained by adding for some i < n to any
cgh in H?(n) all triangles containing a pair vi+j , vi+j+(n−1)/2 for 0 ≤ j ≤ (n− 3)/2 (left diagram in
Figure 3). For n ≥ 4 even, each H ∈ H+(n) consists of all triangles containing the centroid of Ωn in
their interior or on their boundary (right diagram in Figure 3).
Figure 3: Construction of H+(n)
By inspection, each H ∈ H+(n) is M1-free and S1-free. Moreover, if n is odd, then |H| = ·4(n) +
(n− 1)(n− 3)/4 whereas if n is even, then |H| = ·4(n) + n(n− 2)/4. The following theorem answers
Problem 1.
Theorem 1. For all n ≥ 3, if H is an extremal n-vertex M1-free cgh, then H ∈ H′(n) and if H is an
extremal n-vertex M1, S1-free cgh, then H ∈ H+(n). In particular,
ex(n,M1) = ·4(n) +n(n− 3)
2.
ex(n, M1, S1) = ·4(n) +⌊n
2
⌋⌊n− 2
2
⌋.
While we do not determine ex(n, S1) exactly, we prove the following asymptotic result:
Theorem 2. For all n ≥ 3,
·4(n) +⌊n
2
⌋⌊n− 2
2
⌋≤ ex(n, S1) ≤ ·4(n) + n2.
Theorems 1 and 2 show that ex(n,M1) and ex(n, S1) are strictly larger than ex(n, M1, S1, D1)by an additive quadratic term. We do not optimize the upper bound in Theorem 2 in this paper; a
more delicate codegree argument in the proof of Theorem 2 gives ex(n, S1) ≤ ·4(n) + (7/16)n2. We
conjecture that all extremal S1-free n-vertex cghs are in H+(n), and in particular:
Conjecture 1. For all n ≥ 5, ex(n, S1) = ·4(n) + bn/2cb(n− 2)/2c.
A family of triangles with vertices from an n-point set in the plane is intersecting if any two of
the triangles share a point in their interior. Frankl, Holmsen and Kupavskii [8] posed the following
problem:
4
Problem 2. What is the maximum, over all point sets P of size n, of the size of the largest intersecting
family of triangles spanned by P? Is the maximum always at most (1/4 + o(1))(n3
)as n→∞?
Two triangles in the plane form D1 if they have a common edge and the third points of each triangle
are on opposite sides of the line defined by the edge. We solve the above problem exactly as follows:
Theorem 3. Let n ≥ 3. Then the maximum over all sets of n points in the plane, no three collinear,
of the size of a D1-free family of triangles spanned by the points is ·4(n).
If t(n) denotes the maximum in Problem 2, then Theorem 3 shows t(n) = ·4(n) = ex(n,D1), since
the family H ∈ H?(n) with vertex set P = Ωn is intersecting. There are a number of extremal
constructions for Theorem 3: for example, when n is a power of 3, a completely different D1-free
construction to H?(n) with ·4(n) edges is obtained by partitioning Ωn into three intervals of equal
length, adding all triples with one point from each interval, and then repeating this construction within
each interval. Note that Theorem 3 also sharpens the main result of [8].
1.2 Configurations M2 and M3. The extremal function for M3 was determined exactly in [9]
with Jiang and Kostochka:
Theorem 4 ([9]). For all n ≥ 3,
ex(n,M3) =
(n
3
)−(n− 3
3
).
For the sake of completeness, we include the short proof of the theorem from [9]. An extremal M3-free
construction is simply to take all(n−1
2
)triples containing a fixed vertex v0, plus the set of all n(n− 3)
triples which contain a pair of cyclically consecutive vertices of Ωn. It turns out this is not the only
M3-free construction with that many edges: we may remove any triple v0, v2k+1, v2k+3 and add
v2k, v2k+2, v2k+4 when 2k + 4 < n to obtain many different extremal constructions.
An M2-free construction on Ωn is obtained by taking all triples containing a fixed vertex, plus all n
triples of three cyclically consecutive vertices. This construction has(n2
)−2 triples, and we prove that
this construction is extremal:
Theorem 5. For all n ≥ 7,
ex(n,M2) =
(n
2
)− 2.
The restriction n ≥ 7 is necessary in this theorem, since for n = 6, the only copies of M2 on Ω6 are the
triples v0, v1, v3, v0, v2, v3, v0, v1, v4, v0, v3, v4, v0, v2, v5, v0, v3, v5 with their corresponding
complements. As such, removing exactly one pair from each copy of M2 from the complete cgh on
Ω6 gives an M2-free cgh H with 14 =(n2
)− 1 triples. It is likely the case that the star plus the set
of triples of consecutive vertices in Ωn is the unique extremal M2-free example up to isomorphism
for n ≥ 8. For n = 7, we may take all seven cyclically consecutive triples, the edge v1, v3, v6, and
5
all edges which contain v0 besides the edge v0, v4, v5. Similarly, when n = 7, we may also take all
seven cyclically consecutive triples, the edges v1, v3, v6 and v1, v4, v6, and all edges which contain
v0 besides the edges v0, v4, v5 and v0, v2, v3.
1.3 Configurations S2 and S3. For the configuration S3, we are also able to determine the ex-
tremal function exactly when n is even:
Theorem 6. For all even n ≥ 4,
ex(n, S3) =n(n− 2)
2.
For the lower bound ex(n, S3) ≥ n(n− 2)/2 when n is even, let
H0 := v2i−1, v2i, v ∈(
Ωn
3
): 0 ≤ i < n/2.
By inspection, H0 is S3-free and has n(n− 2)/2 edges.
The configuration S2 appears to be the most difficult to handle; we give the following bounds:
Theorem 7. As n→∞, ⌊n2
4
⌋− 1 ≤ ex(n, S2) ≤ 23
64n2 + O(n).
A construction demonstrating the lower bound is to split Ωn into two intervals A and B, and to
take all triples which contain a point from A and a pair of consecutive points in B. We also add
all triples containing three consecutive points in B. This configuration has |A|(|B| − 1) + |B| − 2 =
(|A| + 1)(|B| − 1) − 1 triples and does not contain S2. If |A| = dn/2e − 1 and |B| = bn/2c + 1 then
this configuration has bn2/4c − 1 triples. We believe that this construction should be optimal:
Conjecture 2. For all n ≥ 5, ex(n, S2) = bn2/4c − 1.
1.4 Configuration D2. If G is a convex geometric graph that is a triangulation of a convex polygon,
then the set T (G) of triangles in G form a D2-free cgh. By Euler’s Formula, if |G| = 2k − 3 then
|T (G)| = k − 2. Furthermore, if G1, G2, . . . , GN are edge-disjoint triangulations of polygons with
vertices from Ωn, then H = T (G1) ∪ T (G2) ∪ · · · ∪ T (GN ) is a D2-free cgh on Ωn. If |Gi| = 2k − 3 for
all i ≤ N and all edges of a convex geometric Kn on Ωn are contained in the Gi, then
|H| = k − 2
2k − 3·(n
2
)=
1
2
(n
2
)− 1
4k − 6
(n
2
).
This argument can be used to give an upper bound ex(n,D2) ≤ (1/2 + o(1))(n2
). We will improve
this bound slightly in the theorem below and also give a nontrivial lower bound.
6
Theorem 8. As n→∞,
8
21
(n
2
)+ O(n) ≤ ex(n,D2) ≤ (0.499)
(n
2
).
No strong attempt is made to optimize the proof of the upper bound in this theorem, as we do not
have a conjecture for the asymptotic value of ex(n,D2).
Problem. Determine the asymptotics for ex(n,D2).
1.5 Summary of results and conjectures. We summarise the results and conjectures of this
paper in the following table:
F ex(n, F ) F Bounds on ex(n, F ) Conjecture
·4(n) + n(n−3)2 ·4(n) + [bn2 cb
n−22 c, n
2] ·4(n) + bn2 cbn−2
2 c
(n2
)− 2 [bn2
4 c − 1, 23n2
64 + O(n)] bn2
4 c − 1
(n3
)−(n−3
3
)·4(n) −
n(n−2)2 [ 8
21
(n2
), (0.499)
(n2
)] ?
1.6 Notation and organization Throughout the paper, let Ωn = v0, . . . , vn−1, with a clockwise
cyclic ordering of the vertices. A segment of Ωn is a set of cyclically consecutive vertices of Ωn.
When we write vi < vj < vk, we mean that if we traverse Ωn in clockwise order, we encounter the
vertices in this order. A convex geometric hypergraph or cgh for short is a family H ⊂(
Ωn
3
). We
say that H is F -free if H does not contain F as a subhypergraph. Given H ⊂(
Ωn
3
)and A ⊆ Ωn,
let dH(A) = |e ∈ H : A ⊂ e| be the degree of A in H; we write dH(u, v) when A = u, v and
dH(v) when A = v. For a graph G and v ∈ V (G), let NG(v) = w ∈ V (G) : v, w ∈ G be the
neighborhood of v in G. Let ∂H = u, v : ∃e ∈ H, u, v ⊂ e denote the shadow of H.
7
We employ standard asymptotic notation throughout. For functions f, g : N→ R+, we write f ∼ g if
limn→∞ f(n)/g(n) = 1, f = o(g) if limn→∞ f(n)/g(n) = 0, and f = O(g) if there is c > 0 such that
f(n) ≤ cg(n) for all n ∈ N. If f = O(g) and g = O(f), we write f = Θ(g).
For 1 ≤ i ≤ 7 we will prove Theorem i in Section i + 1.
2 Proof of Theorem 1: two separated triangles, M1
In this section, we use a method similar to that in [8] to prove ex(n,M1) = ·4(n) + n(n− 3)/2 and
ex(n, M1, S1) = ·4(n) + bn/2cb(n− 2)/2c for all n ≥ 3.
2.1 Avoiding M1. Here we prove that if H is an n-vertex extremal M1-free cgh, then H ∈ H′(n).
First let n ≥ 3 be odd. If H ∈ H′(n) then we are done, so we may assume H contains a triangle
T (i, j, k) = vi, vj , vk with vi < vj < vk < vi+(n−1)/2. Moreover, we may assume that among all
such triangles, T (i, j, k) is the triangle where the longest edge vi, vk is as short as possible. Replace
all triangles T (i, j′, k) ∈ H with i < j′ < k with all triangles T (i − 1, k + 1, l) where j and l are on
opposite sides of the edge vi, vk as shown in Figure 4. Since T (i, j, k) and T (i − 1, k + 1, l) form a
copy of M1, T (i − 1, k + 1, l) 6∈ H for all such l. Moreover, since vi < vk < vi+(n−1)/2, the number
of triangles T (i − 1, k + 1, l) that we added is greater than the number of triangles T (i, j, k) that we
deleted. Consequently, this produces a cgh H ′ with |H ′| > |H|. Since H is extremal M1-free, there
exists a copy of M1 in H ′, which must contain a triangle T (i − 1, k + 1, l) ∈ H ′. Since all triangles
T (i − 1, k + 1, l) intersect, the other triangle in the copy of M1 must be T (f, g, h) ∈ H. Since H is
M1-free, T (f, g, h) intersects T (i, j, k), which implies vi ≤ vf < vg < vh ≤ vk and vf , vh 6= vi, vk.However, then the edge vf , vh is shorter than the edge vi, vk, a contradiction.
Figure 4: Replacing triangles
Now let n ≥ 4 be even and let H be an extremal n-vertex M1-free cgh. If H ∈ H′(n) we are done, so
suppose H 6∈ H′(n). If H contains a triangle T (i, j, k) where vi < vj < vk < vi+n/2−1, then we repeat
8
the same proof as in the case n is odd to derive a contradiction. Therefore all triangles in H contain
the centroid or are T (i, j, k) with vi < vj < vk = vi+n/2−1. The pairs vi, vi+n/2−1 for which there
exists such a triangle T (i, j, k) must pairwise intersect (possibly at their endpoints) otherwise we find
a copy of M1 in H. In particular, by definition of Construction 2, H ∈ H′(n). 2
2.2 Avoiding M1 and S1. Here we show that if H is an n-vertex extremal M1, S1-free cgh,
then H ∈ H+(n). First let n ≥ 3 be odd. If every triangle in H contains a pair vi, vi+(n−1)/2or contains the centroid of Ωn, then the set of pairs vi, vi+(n−1)/2 such that there exists a triangle
T (i, j, k) = vi, vj , vk with vi < vj < vk = vi+(n−1)/2 must form a matching, otherwise we find a
copy of S1. We conclude from Construction 3 that H ∈ H+(n). So we may assume there is a triangle
T (i, j0, k) = vi, vj0 , vk in H where vi < vj0 < vk < vi+(n−1)/2. We choose T (i, j0, k) so that the
longest side vi, vk of the triangle T (i, j0, k) is as short as possible. Let Hi,k be obtained by replacing
every triangle T (i, j0, k) with every triangle T (k + 1, l, i) where vk+1 < vl < vi, as shown in Figure
5. Since i < k < i + (n − 1)/2, |Hi,k| > |H| and Hi,k contains M1 or S1. It is not hard to see that
Hi,k cannot contain M1, and any copy of S1 in Hi,k must contain a triangle T (i + 1, j1, k + 1) ∈ H,
where vi+1 < vj1 < vk+1, as in Figure 5. Now vi+1, vk+1 has the same length as vi, vk, so we
may repeat the argument in Hi+1,k+1 to obtain a triangle T (i + 2, j2, k + 2) ∈ H. Eventually we
obtain a triangle T (k, jk−i, 2k − i) ∈ H where vk < vjk−i< v2k−i < vi, which forms S1 together with
Ti,j0,k ∈ H, a contradiction. Now suppose n ≥ 4 is even. If all triangles in H contain the centroid or a
pair vi, vi+n/2, then H ∈ H+(n). Otherwise, there is a triangle T (i, j, k) where vi < vj < vk < vi+n/2
and the same proof as above applies. 2
Figure 5: Replacing triangles
9
3 Proof of Theorem 2: touching triangles, S1
Let H ⊂(
Ωn
3
)be a cgh and let Gi = u,w : vi, u, w ∈ H be the link graph of vi ∈ Ωn. We may
assume |Gi| > 0 otherwise we remove the isolated vertex vi and proceed by induction.
Lemma 3.1. Let H ⊂(
Ωn
3
)be an S1-free cgh and vi ∈ Ωn. Then for some mi 6= i, Gi − vmi is
bipartite with parts Ii and Ji contained in disjoint segments of Ωn.
Proof. Let mi be the maximum integer such that Ii = vi+1, vi+3, . . . , vmi−1 is an independent set
in Gi. If Ji = vmi+1, vmi+2, . . . , vi−1 is not an independent set in Gi, then vj , vk ∈ Gi for some
vj , vk > vmi . By the maximality of mi, vh, vmi ∈ Gi for some h : i + 1 ≤ h ≤ mi − 1 and hence
vi, vh, vmi and vi, vj , vk form S1 in H; a contradiction. 2
Lemma 3.2. Suppose m0 = k and mk = `, and V = v0, vk, v`, and let G = G0−V and G′ = Gk−V .
Then G ∪G′ is bipartite.
Proof. By Lemma 3.1, G and G′ are both bipartite, with parts I = I0\V and J = J0\V for G and
parts I ′ = Ik\V and J ′ = Jk\V for G′. Note I ′ and J both “start” at vk+1 – see Figure 6. If k < ` < n,
then v0, vj , vk ∈ H for some vj with 0 < j < k, by definition of k = m0, and vk, vm, v` ∈ H for
some vm with k < m < `. Now v0, vj , vk and vk, vm, v` form a copy of S1 in H, as shown in the
left diagram in Figure 6, a contradiction. Therefore 0 ≤ ` < k. In this case, J ⊆ I ′. Suppose for
some v ∈ I ∩ I ′ there exist w ∈ NG(v) and w′ ∈ NG′(v) where w 6= w′. Then w ∈ J and w′ ∈ J ′, and
v, w, v0 and v, w′, vk form a copy of S1 in H, a contradiction – see the right diagram in Figure 6.
Defining Y = v ∈ I ∩ I ′ : NG(v) ⊆ J and Y ′ = v ∈ I ∩ I ′ : NG(v) ⊆ J ′, the sets Y ∪ J ′ and Y ′ ∪ Jare the parts of G ∪G′. 2
Figure 6: Proof of Lemma 3.2.
Lemma 3.3. |G ∩G′| ≤ (n− 3)/2.
Proof. We may assume |I| ≤ |J |. Observe that since I and J are independent sets in G, then they are
also independent sets in G∩G′. Suppose there exists v ∈ I with two neighbors u,w ∈ J in G∩G′, where
10
u precedes w in the clockwise ordering on J . Then u, v, vk and v, w, v0 form S1, a contradiction.
So every vertex of I has at most one neighbor in J in G ∩G′, and |G ∩G′| ≤ |I| ≤ (n− 3)/2. 2
Lemma 3.4. Let H ⊂(
Ωn
3
)be an S1-free cgh. Then dH(v0) + dH(vk) ≤ n2/4 + 3n.
Proof. Using dH(u, v) ≤ n− 2 for all u, v ∈ Ωn, we have
dH(v0) + dH(vk) ≤ |G|+ |G′|+ dH(v0, v`) + dH(vk, v`) + 2dH(v0, vk) ≤ |G|+ |G′|+ 4(n− 2).
By Lemma 3.2, G∪G′ is bipartite so |G∪G′| ≤ (n− 3)2/4. Since |G∩G′| ≤ (n− 3)/2 by Lemma 3.3:
|G|+ |G′| = |G ∪G′|+ |G ∩G′| ≤ (n− 3)2
4+
n− 3
2≤ n2
4− n +
3
4.
Adding this to the preceding inequality gives the lemma. 2
We now complete the proof of ex(n, S1) ≤ ·4(n) + n2. This is easily true when n = 3. Suppose
n > 3 and let H be an n-vertex S1-free cgh with ex(n, S1) edges. By Lemma 3.4, H has a vertex v
of degree at most n2/8 + 3n/2. Then
ex(n, S1)− n2
8− 3n
2≤ |H − v| ≤ ex(n− 1, S1).
By induction ex(n − 1, S1) ≤ r(n − 1), where r(n) = r(n − 1) + n2/8 + 3n/2 and r(3) = 1. Solving
this recurrence gives r(n) = n3/24 + 13/16n2 + 37n/48− 39/4. Recalling ·4(n) ≥ n(n2 − 4)/24 for all
n ≥ 3, we obtain for n ≥ 4:
ex(n, S1) ≤ ·4(n) + r(n)− ·4(n) ≤ ·4(n) +13n2
16+
45n
48− 39
4< ·4(n) + n2.
This completes the proof. 2
4 Proof of Theorem 3: tangent triangles, D1
Let T (n) be the maximum number of directed triangles in an n-vertex tournament. It was shown
by Moon [20] (see also pages 42–44 in Erdos and Spencer [6]) that T (n) = ·4(n) for n ≥ 3. To see
this, every tournament with n vertices of outdegrees d1, . . . , dn has exactly(n3
)−∑n
i=1
(di2
)directed
triangles. This is maximized when the di are as equal as possible. If n is odd, then all di = (n− 1)/2
while if n is even, half of the di are (n− 2)/2 and the other half are n/2. Moreover, tournaments with
these outdegrees can easily be constructed, and a short calculation gives the required
T (n) = ·4(n).
We will prove that t(n) ≤ T (n), which gives Theorem 3. Let P be a set of n points in the plane with
no three collinear, and let H be a D1-free family of triangles on P . For every pair vi, vj ∈ ∂H, there
11
is an edge vi, vj , vk ∈ H such that either vi < vk < vj or vi < vj < vk. Note that both possibilities
cannot happen simultaneously since that would yield a copy of D1. If vi < vk < vj then we orient
vi, vj as (vi, vj) and if vi < vj < vk then we orient vi, vj as (vj , vi). This yields an orientation
of ∂H. The main observation is that there is no edge vi, vj , vk ∈ H such that vi, vj , vk is not a
directed triangle in the orientation. Indeed, if we have (vi, vj) and (vi, vk), where vi < vj < vk, then
there is vl with vi < vl < vj and triangle vi, vl, vj. Together with vi, vj , vk we obtain a copy of
D1. We conclude that |H| is at most the number of directed triangles in an orientation of a subgraph
of Kn which is at most T (n) as required.
5 Proof of Theorem 4: crossing triangles, M3
For the proof of Theorem 4 for M3, it is useful to consider ordered hypergraphs: the vertex set is
Ωn = v0, v1, . . . , vn−1 with the ordering v0 < v1 < · · · < vn−1. Let ex→(n,M3) denote the maximum
number of triples in an ordered hypergraph not containing triples vi, vj , vk and vi′ , vj′ , vk′ with
vi < vi′ < vj < vj′ < vk < vk′ – this is the ordered analog of M3. The following theorem implies
Theorem 4, since ex(n,M3) = ex→(n,M3):
Theorem 9. Let n ≥ 7. Then ex→(n,M3) =(n3
)−(n−3
3
).
Proof. Let H be an M3-free ordered triple system with n vertices. Let H1 consists of all e ∈ H
with v0, v1 ∈ e, and let H2 consists of all e ∈ H with v0 ∈ e, v1 6∈ e and e − v0 + v1 ∈ H. Let
H3 be obtained from H\(H1 ∪ H2) by merging the vertices v0 and v1. Note that H3 is a 3-cgh
with n − 1 vertices. Clearly, |H1| ≤ n − 2. We may form an ordered graph from H2 by considering
G = u, v : v0, u, v ∈ H2 – this is the link graph of v0 with vertex set v2, v3, . . . , vn−1 with the
natural ordering. If two edges of G cross – say u, v, w, x ∈ G with u < w < v < x, then the triples
u, v, v1 and w, x, v0 are in H2, and form a copy of M3, a contradiction. Therefore no two edges of
G cross, which implies G is an outerplane graph with n − 2 vertices. Consequently |G| ≤ 2n − 7, by
Euler’s Formula. Finally, it is also straightforward to check H3 is M3-free, so by induction,
|H| = |H1|+ |H2|+ |H3| ≤ (n− 2) + (2n− 7) +
(n− 1
3
)−(n− 4
3
)=
(n
3
)−(n− 3
3
).
This completes the proof of Theorem 9. 2
6 Proof of Theorem 5: stabbing triangles, M2
We prove by induction on n that ex(n,M2) =(n2
)− 2 for n ≥ 7. When n = 7, since cyclically
consecutive triples vi, vi+1, vi+2 are never in M2, we may assume these seven edges are in any M2-
free cgh. For the remaining twenty-eight triples, we create a graph with vertex sets consisting of these
triples and form an edge if two of the triples form a copy of M2. A computer aided calculation [26]
12
then yields this graph has independence number 12 and hence ex(7,M2) = 12 + 7 =(
72
)− 2.
For the induction step, we plan to find two consecutive u, v ∈ Ωn with degree at most three and whose
common link graph Gu ∩ Gv has at most n − 3 edges. Let H be a maximal M2-free cgh on Ωn, and
H ′ ⊂ H be the cgh after removing all consecutive triples vi, vi+1, vi+2. Let d(vi, vj) be length of
the path on the perimeter of the polygon starting with vi and moving clockwise to vj . For an edge
e = vi, vi+1, vk ∈ H ′ – we only consider such edges – let `(e) = mind(vi+1, vk), d(vk, vi).
Lemma 6.1. Let H ⊂(
Ωn
3
)be a maximal M2-free cgh and H ′ be as above. Then
(1) For consecutive u, v ∈ Ωn, |Gu ∩Gv| ≤ n− 3 with equality only if Gu ∩Gv is a star.
(2) There exists vi ∈ Ωn such that the degree of vi, vi+1 is at most three in H.
Proof. We first prove (1) by showing Gu,v := Gu ∩ Gv does not contain a pair of disjoint edges If
w, x, y, z are disjoint edges in Gu,v, and v < w < x < y < z < u < v or v < w < y < z < x < u < v
– this means that w, x, y, z do not cross – then u,w, x, v, y, z form M2. If on the other hand
v < w < y < x < z < u < v – this means w, x, y, z do cross – then u, y, z, v, w, x form M2. So
Gu,v has no pair of consecutive edges. It is a standard fact that the unique extremal graphs with at
least four vertices and no pair of disjoint edges are stars, and therefore Gu,v has at most n− 3 edges.
For (2), seeking a contradiction, suppose every pair of consecutive vertices has degree at least four in
H and hence degree at least two in H ′. We first show there exists e ∈ H ′ with `(e) ≥ 3. If not, then
vi, vi+1, vi+3 ∈ H ′ and vi−2, vi, vi+1 ∈ H ′ for all i and there are no other edges in H ′. However, then
v0, v1, v3 ∈ H ′ and v2, v4, v5 ∈ H ′ form M2, a contradiction. So there exists e ∈ H ′ with `(e) ≥ 3.
From all e ∈ H ′ with `(e) ≥ 3, pick e so that `(e) = j ≥ 3 is a minimum. Suppose e = v0, v1, vj+1,so `(e) = d(v1, vj+1) (the proof for e of the form vn−j , v0, v1 with `(e) = j = d(vn−j , v0) ≥ 3 will be
symmetric). Then the pair vj−1, vj has degree at least two in H ′ so there are edges f = vh, vj−1, vjand g = vk, vj−1, vj in H ′. If j + 1 < k ≤ n − 1 or j + 1 < h ≤ n − 1, then f and e or g and e
respectively form M2, a contradiction. So 0 ≤ h, k ≤ j − 3, recalling vj−2, vj−1, vj 6∈ H ′. Now
`(f) = d(vh, vj−1) > d(vk, vj−1) ≥ 2
and so `(f) ≥ 3. On the other hand, since 0 ≤ h < j − 1,
`(f) = d(vh, vj−1) < d(v0, vj) = `(e)
contradicting the choice of e. This final contradiction proves (2). 2
Let vi, vi+1 have degree at most three in H, as guaranteed by Lemma 6.1 part (2). We con-
tract the pair vi, vi+1 to a vertex w to get a cgh H0 with n − 1 vertices. Let G = u, v :
u, v, vi, u, v, vi+1 ∈ H be the common link graph of vi and vi+1.
Lemma 6.2. Let G be the common link graph of vi and vi+1. Then |G| ≤ n− 4.
Proof. If neither of vi−1, vi, vi+2 or vi−1, vi+1, vi+2 is in H, then vi−1, w, vi+2 6∈ H0 and |G| ≤
13
n− 4 follows from Lemma 6.1 part (1). So we assume vi−1, vi, vi+2 ∈ H or vi−1, vi+1, vi+2 ∈ H.
Case 1. vi−1, vi, vi+2 ∈ H. Suppose G is a star with n−3 edges, with center vk. If vk /∈ vi−1, vi+2,then letting vj /∈ vk, vi−1, vi, vi+1, vi+2, it follows that vi, vj , vk and vi−1, vi+1, vi+2 form a copy
of M2. Hence, we may assume that vk = vi−1 or vk = vi+2. Both of these cases are similar, so consider
only the case vk = vi+2. We may assume that vi+3, vi+4 has degree at least three. Then there is at
least one triple which contains vi+3, vi+4 of the form v, vi+3, vi+4. If v ∈ Ωn and vi+4 < v < vi+1,
then v, vi+3, vi+4 and vi+1, vi+2, vi+5 form M2. If v = vi+1, then v, vi+3, vi+4 and vi−1, vi, vi+2form M2. So G is not a star with n− 3 edges, and Lemma 6.1 part (1) gives |G| ≤ n− 4.
Case 2. vi−1, vi+1, vi+2 ∈ H. In this case, a symmetric argument to that used for vi−1, vi, vi+2 ∈H applies by reversing the orientation of Ωn. 2
To complete the proof of |H| ≤(n2
)− 2, we note by inspection that H0 is also M2-free. By induction,
|H0| ≤(n−1
2
)− 2. By Lemma 6.2, and recalling dH(vi, vi+1) ≤ 3,
|H| = |H0|+ |G|+ dH(vi, vi+1) ≤(n− 1
2
)− 2 + n− 4 + 3 =
(n
2
)− 2.
This proves Theorem 5. 2
7 Proof of Theorem 6: crossing triangles with a common vertex, S3
Let H ⊂(
Ωn
3
)be a S3-free cgh and Gi be the link graph of vi in H. Let G′i comprise the edges of Gi
which consist of two consecutive vertices in Ωn, and let G′′i = Gi\G′i.
Lemma 7.1. Let H ⊂(
Ωn
3
)be a S3-free cgh. For 0 ≤ i ≤ n− 1, |G′′i | ≤ n− 3.
Proof. The graph G′′i has no pair of crossing edges since H is S3-free. We may assume i = 0. If we
add to G′′i all n− 2 edges vj , vj+1 for 1 ≤ j < n, we obtain an ordered graph G with n− 1 vertices,
and |G| ≤ 2(n− 1)− 3 = 2n− 5. Removing the n− 2 added edges gives |G′′i | ≤ n− 3. 2
Lemma 7.2. Let H ⊂(
Ωn
3
)be a S3-free cgh. For each i with 0 ≤ i ≤ n/2− 1, |G′2i|+ |G′2i+1| ≤ n.
Proof. We may assume i = 0. Let G denote the multigraph obtained by superimposing the graphs G′0and G′1, so |G| = |G′0| + |G′1|. Each component C of G is a path P with some edges of multiplicity
two. If vj−1, vj ∈ P ∩ G′0, then vj , vj+1 6∈ P ∩ G′1, otherwise v0, vj , vj+1, v1, vj−1, vj form
S3 ⊂ H as in Figure 7, a contradiction. If all edges of P are from G′1 only, then |C| = |P | =
|V (C)| − 1. Otherwise, let vj , vj+1 be the first edge of P in G′0 in the clockwise direction. Then
all edges of P preceding vj , vj+1 are in G′1 only, and all edges of P after vj , vj+1 are in G′0only, whereas vj , vj+1 might be in both G′0 and in G′1. Therefore at most one edge of P has
multiplicity two, and |C| ≤ |P |+ 1 = |V (C)|. If C1, C2, . . . , Cr are the components of G, we conclude
|G| = |C1|+ |C2|+ · · ·+ |Cr| ≤ |V (C1)|+ |V (C2)|+ · · ·+ |V (Cr)| = |V (G)| = n. 2
14
Figure 7
We now complete the proof of ex(n, S3) ≤ n(n− 2)/2, using the following identity:
3|H| =n−1∑i=0
(|G′i|+ |G′′i |) =
n/2−1∑i=0
(|G′2i|+ |G′2i+1|) +n−1∑i=0
|G′′i |.
We apply Lemmas 7.1 and 7.2 to each term in the sums to obtain:
3|H| ≤n/2−1∑i=0
n +n−1∑i=0
(n− 3) =1
2n2 + n(n− 3) =
3
2n(n− 2). 2
8 Proof of Theorem 7: touching triangles with parallel sides, S2
Let H ⊂(
Ωn
3
)be an S2-free cgh. We are going to show |H| ≤ 23n2/64 + O(n). Consider an edge
e = vi, vj , vk ∈ H where vi < vj < vk. We call the pair vi, vj good for e if there does not exists a
k′ such that vj < vk′ < vk and vi, vj , vk′ ∈ H, and bad otherwise.
Lemma 8.1. Let H ⊂(
Ωn
3
)be an S2-free cgh. Then
(1) Every edge of H contains at least two good pairs.
(2) Every pair in ∂H is good for either one or two edges of H.
Proof. We first prove (1). Suppose e = vi, vj , vk ∈ H and vi, vj and vj , vk are bad. Then there
exist k′ : vj < vk′ < vk and i′ : vk < vi′ < vi such that vi, vj , vk′, vj , vk, vi′ ∈ H. However, the
edges vi′ , vj , vk and vi, vj , vk′ form configuration S2, a contradiction.
For (2), given vi, vj ∈ ∂H, consider an edge vi, vj , vk with vi < vj < vk and vk as close as possible
to vj ; this determines vk uniquely. Similarly, for vi, vj ∈ ∂H, consider an edge vi, vj , vk with
vi < vk < vj with vk as close as possible to vi; this too determines vk uniquely. Therefore each pair in
∂H is good for either one of two edges of H. 2
15
Color a pair in ∂H blue if it is good for exactly one edge in H, and red if it is good for exactly two
edges in H. Let R be the number of red pairs and B the number of blue pairs – for a red pair u, v,there exist vertices w, x ∈ Ωn on opposite sides of u, v such that u, v, w ∈ H and u, v, x ∈ H. If
we map an edge e ∈ H to the pairs in e that are good for e, then each red pair is counted twice and
each blue pair is counted once. On the other hand, each edge of H contains at least two good pairs,
by Lemma 8.1, so 2|H| ≤ 2R + B. In particular,
|H| ≤ R + B/2 ≤ R + B = |∂H|.
Lemma 8.2. If vi, vj, vjvk and vk, vi are red pairs, then vi, vj , vk ∈ H.
Proof. Suppose vi, vj , vk 6∈ H and vi < vj < vk. Then by definition there exists k′ 6= k such that
vi, vj , vk′ ∈ H and vj < vk′ < vi. We consider two cases.
Case 1. vj < vk′ < vk. There exists i′ 6= i such that vi′ , vj , vk ∈ H and vk < vi′ < vj . We observe
vi < vi′ < vj , otherwise vj , vk′ and vi, vi′ are non-crossing, and vi, vj , vk′ and vi′ , vj , vk form
S2 in H. Now there exists j′ 6= j such that vi, vj′ , vk ∈ H and vi < vj′ < vk. If vi < vj′ < vj ,
then the pairs vj′ , vk and vj , vk′ are non-crossing, and vi, vj , vk′ and vi, vj , vk′ form S2. If
vj < vj′ < vk, then vi′ , vj and vi, vj are “parallel”, and vi′ , vj , vk and vi, vj , vk′ form S2 in H.
Case 2. vk < vk′ < vi. Consider the reverse ordering of Ωn and apply the proof of Case 1. 2
By Lemma 8.2, every triangle of red pairs is an edge of H, so there are at most |H| ≤ |∂H| ≤(n2
)such triangles. In particular, the number of red pairs is at most n2/4 + O(n) (one could use a precise
result by Lovasz-Simonovits [18] to deduce this). Therefore
2|H| ≤ 2R + B ≤ 3n2
4+ O(n).
To improve this bound to the desired |H| ≤ 23n2/64 + O(n), we may assume n is odd and partition
the complete graph on Ωn into planar matchings M1,M2, . . . ,Mn where Mi = vj , vk : j + k ≡ i
mod n. Then there exists i ≤ n such that at least R/n pairs in M = Mi are red. For each pair of
red pairs, say u, v and w, x, where u < w < x < v < u, there exist triples u, v, y, w, x, z ∈ H
where u < w < z < x < v < y < u. Now by inspection, the pair y, z cannot be contained in any
edge of H without creating configuration S2 – see Figure 8. Furthermore, if u′, v′, w′, x′ ∈ M ,
then u′, v′, y and w′, x′, z cannot both be edges of H without creating S2. Therefore for each pair
u, v, w, x of red edges of M , we may associate a unique pair y, z which is not contained in
any edge of H. Consequently
2|H| ≤ 2R + B ≤ 2R +
(n
2
)−(R/n
2
)−R ≤ R +
n2
2− (R− n)2
2n2.
Since R ≤ n2/4 + O(n), this implies |H| ≤ 23n2/64 + O(n), as required. 2
16
Figure 8
9 Proof of Theorem 8: triangles sharing a side, D2
For the upper bound on ex(n,D2), let H ⊂(
Ωn
3
)be a D2-free cgh. Make a graph G with vertex set
H: two triangles of H are joined by an edge of G if they share a side. Consider the partition of H
generated by the components C1, C2, . . . Ct of G, where |Ci| = ki. Viewing each Ci as a hypergraph
of triangles, each ∂Ci forms a triangulation of a convex (ki + 2)-gon with 2ki + 1 diagonals and sides.
Now let k = 31. Let C1, . . . , Cm be the components with ki ≥ (2k − 3), and let e =∑m
i=1 |∂Ci|. For
these components we use the bound ki < (1/2)(2ki + 1) = (1/2)|∂Ci|. This yields∑
i≤m ki < (1/2)e.
For the rest of the components ki ≤ 2k − 4 and we use the bound
ki =
(ki
2ki + 1
)(2ki + 1) ≤ 2k − 4
4k − 7|∂Ci|.
Summarizing, we have
|H| =t∑
i=1
ki <e
2+
((n
2
)− e
)2k − 4
4k − 7=
e
2(4k − 7)+
n2
4
(1− 1
4k − 7
)+ O(n).
Next we give a good upper bound for e. Call an edge in ∂H long if its length exceeds n/(k + 1) and
short otherwise. Each Ci has at most k long sides, so it has at least ki + 2−k short sides. Let s be the
number of short sides of Ci. We have |∂Ci| = 2ki + 1 ≤ 4s whenever ki is large. There are n2/(k + 1)
sides of the n-gon with length at most n/(k + 1), and each can appear at most twice as a side of a
triangle from H, so e < 8× n2/(k + 1). Now k = 31 yields an upper bound
|H| ≤(1
4− 1
936+ o(1)
)n2.
For the lower bound on ex(n,D2), start with an S(n, 7, 2) design – Wilson [28] proved these exist
whenever n is large enough and satisfies the requisite divisibility conditions i.e.(n2
)/(
72
)is an integer,
17
and n ≡ 1 mod 6. The construction is as follows: decompose the Kn into(n2
)/(
72
)complete K7’s.
Each corresponds to a convex 7-gon with vertex set V = w1, w2, . . . , w7. Decompose each K7 into
a Fano configuration such that w1, w2, w3, w3, w4, w5, and w5, w6, w1 are triangles. Then the
triangle w1, w3, w5 can be added and ex(n,D2) ≥ (8/21)(n2
). 2
10 Concluding Remarks
• In the cases of M1,M2,M3, S3 and D1, we obtained exact results for the extremal functions. Our
proofs for the first four, with a little more work, should give a characterization of the extremal examples
as well. Note that for M2, one requires n ≥ 8 for the extremal configuration to be unique, as verified
by computer. It seems likely that all extremal S1-free configurations can be determined, but for D1
this appears much more difficult, since very different potentially extremal configurations exist.
• In this paper, we considered the ordered and convex geometric configurations consisting of two
triples. One may consider analogous problems for r-tuples: for instance, how many edges can a
convex geometric n-vertex r-graph have if it does not contain two hyperedges which are geometrically
disjoint as r-gons (this is the r-uniform analog of M1)? This problem was posed explicitly by Frankl,
Holmsen and Kupavskii [8]:
Problem 3. Find analogues of our results for other classes of sets such as convex r-gons in R2.
We solve this problem in a similar way to the proof of Theorem 1: for instance if we forbid in an r-cgh
H two r-gons whose interiors have no point in common, we consider any r-gon vi1 , vi2 , . . . , vir in
H with vi1 < vi2 < · · · < vir < vi1 and where the longest side vi1 , vir is as short as possible, and
replace all such r-gons with vi1 , vj2 , . . . , vjr−1 , vir where vir < vj2 < vj3 < · · · < vjr−1 < vi1 . Since
the number of choices of j2, j3, . . . , jr−1 is always at least the number of choices of i2, i3, . . . , ir−1, this
new r-cgh H ′ has |H ′| ≥ |H|. So we repeat until H ′ consists of all r-gons containing the centroid of Ωn
when n is odd, or n is even and H ′ consists of all r-gons containing the centroid plus for each diameter
we add all r-gons on one side of this diameter. In particular, the extremal function is 21−r(nr
)+O(nr−1)
for each r ≥ 3.
• Since there are many other possible configurations of two r-gons, or two ordered r-tuples, we did not
discuss these problems in this paper. Some special cases were studied in [9]: for instance, if F consists
of two r-tuples u1, u2, . . . , ur and v1, v2, . . . , vr where u1 < v1 < u2 < v2 < · · · < ur < vr < u1,
then it was shown in [9] that for n > r > 1,
ex(n, F ) =
(n
r
)−(
n
r − 2
).
• A hypergraph H is linear if for distinct hyperedges e, f ∈ E(H), |e∩ f | ≤ 1. The extremal functions
for the configurations in this paper in the context of linear hypergraphs were determined in [1] up
18
to constant factors for all the configurations except S2. Specifically, if ex∗(n, F ) is the maximum
number of triples in an n-vertex F -free linear cgh, then Aronov, Dujmovic, Morin, Ooms and da
Silveira [1] proved ex∗(n,M2) = Θ(n), whereas if F ∈ M1,M3, S1, S3, ex∗(n, F ) = Θ(n2). It would
be interesting to determine the exact extremal functions in each case. The problem of determining
ex∗(n, S2) appears to be very difficult, as it is connected to monotone matrices, tripod packing, and
2-comparable sets – see Aronov, Dujmovic, Morin, Ooms and da Silveira [1] for details. The best
bounds are ex∗(n, S2) = Ω(n1.546) due to Gowers and Long [12] and ex∗(n, S2) = n2/ exp(Ω(log∗ n))
due to the best bounds on the removal lemma by Fox [7].
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