Monochromatic triangles in two-colored plane

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by  Vit Jelinek

Monochromatic triangles in two-colored plane
7
0
0
2 V´ıt Jel´ınek1 Jan Kynˇcl2 Rudolf Stolaˇr2 Tom´aˇs Valla2

n
a
J 1 CharlesUniversity,FacultyofMathematics andPhysics,

1 InstituteforTheoretical ComputerScience(ITI)∗
3 Malostransk´ena´m.2/25,11800,Prague,CzechRepublic
[email protected]

]
O 2 CharlesUniversity,FacultyofMathematics andPhysics,
DepartmentofAppliedMathematics (KAM)†
C
Malostransk´ena´m.2/25,11800,Prague,CzechRepublic
.
h [email protected], [email protected], [email protected]
t
a Abstract
m
We prove that for any partition of the plane into a closed set C and
[
an open set O and for any configuration T of three points, there is a

1 translated and rotated copy of T contained in C or in O.
v Apartfromthat,weconsiderpartitionsoftheplaneintotwosetswhose
0 commonboundaryisaunionofpiecewiselinearcurves. Weshowthatfor
4 anysuchpartitionandanyconfigurationT whichisavertexsetofanon-
9 equilateral triangle there is a copy of T contained in the interior of one
1
of thetwo partition classes. Furthermore,we give thecharacterization of
0
these“polygonal”partitionsthatavoidcopiesofagivenequilateraltriple.
7
0 These results support a conjecture of Erd˝os, Graham, Montgomery,
/ Rothschild, Spencer and Straus, which states that every two-coloring of
h
the plane contains a monochromatic copy of any nonequilateral triple of
t
a points; on theotherhand,wedisproveastronger conjecturebythesame
m authors, by providing non-trivial examples of two-colorings that avoid a
given equilateral triple.
:
v
i
X
1 Introduction
r
a
EuclideanRamseytheory addressesthe problems ofthe followingkind: assume
that a finite configuration X of points is given; for what values of c and d is it
true that every coloring of the d-dimensional Euclidean space by c colors con-
tains a monochromatic congruent copy of X? The first systematic treatise on
this theory appears in 1973 in a series of papers [2, 3, 4] by Erdo˝s, Graham,
Montgomery, Rothschild, Spencer and Straus. Since that time, many strong
∗ITIissupportedbyproject1M0021620808 oftheCzechMinistryofEducation
†KAMissupportedbyprojectMSM0021620838 oftheCzechMinistryofEducation
1
results have been obtained in this field, often related to high-dimensional con-
figurations (see, e.g., [5, 7, 8, 9] or the survey [6]); however, there are basic
‘low-dimensional’ problems that remain open.
Inthispaper,weconsiderthespecialcasewhend=2,c=2and X =3;in
| |
other words,we study the configurationsof three points in the Euclideanplane
coloredbytwocolors. Weusethetermtriangle torefertoanysetofthreepoints,
including collinear triples of points, which we call degenerate triangles. An
(a,b,c)-triangleisatrianglewhoseedges,inanti-clockwiseorder,haverespective
lengths a, b and c. A (1,1,1)-triangle is also called a unit triangle.
We saythat aset ofpoints X R2 is a copy ofa setofpoints Y R2, if X
⊆ ⊆
can be obtained from Y by translations and rotations in the plane. A coloring
is a partition of R2 into two sets B and W. The elements of B and W are
called black points and white points, respectively. We use the term boundary
of χ to refer to the common boundary of the sets B and W. Given a coloring
χ = (B,W), we say that a set of points X is monochromatic, if X B or
X W. ⊆
⊆
We say that a coloring χ contains a triangle T, if there exists a monochro-
matic set T which is a copy of T; otherwise, we say that χ avoids T.
′
A coloring that avoids the unit triangle is easy to obtain: consider a color-
ing χ that partitions the plane into alternating half-open strips of width √3;
∗ 2
formally, a point (x,y) is black if an only if n√3 < y n+ 1 √3 for some
≤ 2
integern. Itcanbeeasilycheckedthatχ∗ avoidstheunitt(cid:0)riangle(cid:1). Wecaneven
change the color of some of the points on the boundaries of the strips without
creating any monochromatic unit triangle. Erdo˝s et al. [4, Conjecture 1] have
conjecturedthatthisisessentiallytheonlyexampleofcoloringsavoidingagiven
triangle:
Conjecture 1.1(Erd˝osetal.[4]). Forevery triangleT andevery coloring χ, if
χ avoids T, then T is an equilateral (l,l,l)-triangle and χ is equal to an l-times
scaled copy of the coloring χ defined above, up to possible modifications of the
∗
colors of the points on the boundary of the strips.
In Section 3 of this paper, we present a counterexample to this conjecture,
anddefine ageneralclassofcolorings(whichincludes χ asa specialcase)that
∗
avoid the unit triangle.
Ontheotherhand,thefollowingconjecturebyErdo˝setal. [4,Conjecture3]
remains open:
Conjecture 1.2 (Erd˝osetal.[4]). Every coloring χ contains every nonequilat-
eral triangle T.
In the past, it has been shown that Conjecture 1.2 holds for special types
of triangles T (see, e.g., [4, 6, 10]). Our approach is different: we prove that
the conjecture is valid for a restricted class of colorings χ and arbitrary T. In
Section 2, we show that every coloring that partitions R2 into a closed set and
anopensetcontainseverytriangleT. Then,inSection3,weconsiderpolygonal
colorings, whose boundary is a union of piecewise linear curves (see page 6 for
2
C′
A′ B′
D′
D
c
b
B a A
C
Figure 1: The illustration of the proof of Lemma 1.3
theprecisedefinition). WeshowthatConjecture1.2holdsforthepolygonalcol-
orings,buttherearepolygonalcounterexamplestothe strongerConjecture1.1.
In fact, we are able to characterize all these polygonal counterexamples.
The following lemma from [4] offers a useful insight into the topic of mono-
chromatic triangles in two-coloredplane:
Lemma 1.3. Let χ be a coloring of the plane. The following holds:
(i) If χ contains an (a,a,a)-triangle for some a > 0, then χ contains an
(a,b,c)-triangle, for every b,c > 0 such that a,b,c satisfy the (possibly
degenerate) triangle inequality.
(ii) If χ contains an (a,b,c)-triangle, then χ contains an (x,x,x)-triangle for
some x a,b,c .
∈{ }
Proof. The essence of the proof is the configuration in Figure 1. The configu-
rationconsists of two (a,a,a)-triangles ABC and AB C , two (b,b,b)-triangles
′ ′ ′
ADB and AD B and two (c,c,c)-triangles BDC and B D C. To prove the
′ ′ ′ ′ ′ ′
first part of the lemma, assume, for a given χ, that there is a monochromatic
(a,a,a)-triangle ABC, and choose arbitraryb and c satisfying triangle inequal-
ity with a. Assume that A, B and C are all black. Furthermore, assume for
contradiction that no (a,b,c)-triangle is monochromatic. Considering the con-
figuration in Fig. 1, we deduce that the points B , D and D are all white,
′ ′
otherwise one of the (a,b,c)-triangles BAD, CAB and CBD would be mono-
′ ′
chromatic. Then, A is black, due to B AD , and C is white, due to C AB.
′ ′ ′ ′ ′ ′ ′
It follows that C B D is monochromatic, a contradiction.
′ ′
The second part is proved by an analogous argument: assume that BAD
is an all-white monochromatic triangle and that the statement does not hold.
Then B , C and C are all black, due to ADB , ABC and BDC . A is white,
′ ′ ′ ′ ′
due to AB C ; D is black, due to AD B, and B D C is monochromatic.
′ ′ ′ ′ ′ ′ ′ ′
3
This concludes the proof.
From Lemma 1.3, we obtain directly the following facts:
Corollary 1.4. For every coloring χ the following holds:
(i) χcontainseverytriangleifandonlyifχcontainseveryequilateraltriangle.
(ii) χ contains every non-equilateral triangle if and only if there is an a0 >0
such that χ contains the equilateral (a,a,a)-triangle for all values of a>0
different from a0.
(iii) χ contains an (a,b,c)-triangle if and only if χ contains a (b,a,c)-triangle.
2 Coloring by closed and open sets
The aim of this section is to prove the following result:
Theorem 2.1. Let χ = (B,W) be a coloring such that B is closed and W is
open. Then χ contains every triangle T.
By Corollary 1.4, it suffices to prove Theorem 2.1 for the case when T is
an arbitrary equilateral triangle. Moreover, since scaling does not affect the
topologicalpropertiesofBandW,weonly needto considerthe casewhenT is
the unit triangle. Before stating the proof, we introduce a definition and prove
an auxiliary result.
Definition 2.2. Let ε > 0. An (a,b,c)-triangle whose edge-lengths satisfy
1 ε a,b,c 1+ε is called an ε-almost unit triangle.
− ≤ ≤
Supposethatanorthogonalcoordinatesystemisgivenintheplane. Fora>
0, let Q(a) be the closed square with vertices (a,a),( a,a),( a, a),(a, a).
− − − −
Proposition 2.3. Let Q(3) = B W be a decomposition of the square Q(3)
∪
intotwo disjoint sets suchthat thereis no monochromatic unit triangle in Q(3).
Then for every ε>0 both B and W contain an ε-almost unit triangle.
Proof. Let ε be a given positive number. Assume that we are given a partition
B W=Q(3)suchthatQ(3)doesnotcontainanymonochromaticunittriangle.
Fo∪r contradiction, assume that one of the classes, wlog the class B, does not
contain any ε-almost unit triangle.
There is a white point S and a black point R in Q(1) such that R S <ε
(otherwise the whole square Q(1) would be monochromatic). Let C|be−the|unit
circle centered at S. For every α R, let K(α) denote the point of C with
∈
coordinates (x +cos(α),y +sin(α)), where (x ,y ) are the coordinates of S.
S S S S
NotethatthedistancebetweenRandanypointonCisalwaysintheinterval
(1 ε,1+ε);thus,foreveryα,thepointsK(α)andK(α+π)musthavedifferent
− 3
colors, otherwise they would form a monochromatic white unit triangle with S
or a monochromatic black ε-almost unit triangle with R.
4
π
K(γ) K(α0+ 3)
K(β)
K(β2)
A K(α0)
K(β1)
S
R
Figure 2: Illustration of the proof of Proposition 2.3
Let K(α0) be a white point, then K(α0 + π3) is black (see Fig. 2). Note
that for every α ∈ (α0−ε,α0+ε) the distance between K(α) and K(α0+ π3)
is in the interval (1 ε,1+ε), so the whole arc K(α);α (α0 ε,α0 +ε)
is white. Let A = −K(α);α (β1,β2) be the m{aximal o∈pen w−hite arc of C}
cbT(γolhanectrkaeβ.in)aBilnysgo(dteπehxfieinsptiεtso,iioγnπnt){∈,oKsf(o(βAα2t,0h+)te.hedπ3Tr∈iehs−teeanx2εni,tcsβhets2eb+βwe}th∈wπ3oe)l(eeβsn2au,rctβchh2{et+Khbal(2εaαt)c)kKs;uαp(cγoh∈)int(ithsβsa1bKt+laK(cπ3βk(,)β.β)a2Bni+sudtbπ3Klt)ah}(ceγkins).
− ∈ 3 − 3
is in the interval (1 ε,1), hence the three points R,K(β),K(γ) form a black
−
ε-almost unit triangle—a contradiction.
We are now ready to prove the main result of this section.
Proof of Theorem 2.1. Letχ=(B,W) be acoloring,with Bclosed. By Corol-
lary 1.4, it is sufficient to show that χ contains the unit triangle. Assume,
for contradiction, that this is not the case. Let B0 = Q(3) B and let
W0 = Q(3) W. Clearly, neither B0 nor W0 contain the unit∩triangle, so
∩
by Proposition 2.3, both these sets contain ε-almost unit triangles for every
ε > 0. In particular, the set B0 contains, for every n ∈ N, a n1-almost unit
triangle X Y Z .
n n n
SinceB0isacompactset,thesetB30 =B0 B0 B0iscompactaswell. The
sequence (X ,Y ,Z );n N is an infinite s×equen×ce of points in B3, so there
n n n 0
exists a co{nvergent subseq∈uen}ce (X ,Y ,Z );k N . Let (X,Y,Z) B3
be its limit. Then X,Y,Z B{ arneklimnikts onfkthe∈sequ}ences X ;k ∈N 0,
Y ;k N , and Z ;k ∈N , respectively. The Euclidean{ dnisktanc∈e is}a
{ nk ∈ } { nk ∈ }
continuous function of two variables, so X Y = lim X Y = 1,
similarly Y Z = Z X = 1. Thus,| X−,Y,|Z is akb→l∞ack| unnkit−trinakn|gle in
| − | | − | { }
Q(3), which is a contradiction.
5
3 Polygonal colorings
Throughout this section, C(A) denotes the unit circle with center A, and D(A)
denotes the closed unit disc with center A.
In this section, we consider polygonal colorings of the plane, defined as fol-
lows:
Definition 3.1. A coloring χ = (B,W) is said to be polygonal, if it satisfies
the following conditions (see an example in Fig. 3):
Figure 3: Example of a polygonal coloring
Each of the two sets B and W is contained in the closure of its interior.
•
The boundary of χ (denoted by ∆) is a union of straight line segments
•
(called boundary segments). Two boundary segments may only intersect
at their endpoints. We allow these segments to be unbounded, i.e., a
boundary segment may in fact be a half-line or a line. An endpoint of
a boundary segment is called a boundary vertex. We may assume that
if exactly two boundary segments meet at a boundary vertex, then the
two segments do not form a straight angle, because otherwise they could
be replaced with a single boundary segment. Note that with this condi-
tion, the boundary segments and boundary vertices of χ are determined
uniquely.
Every bounded region of the plane is intersected by only finitely many
•
boundary segments (which implies that every bounded region contains
only finitely many boundary vertices).
Note that these conditions imply that a sufficiently small disc around an
interior point of a boundary segment is separated by the boundary segment
into two halves, one of which is colored black and the other white. Note also
thatwemakenoassumptionsaboutthecolorsofthepointsontheboundary∆.
We say that a coloring χ is a twin of a coloring χ if the two colorings have
′
the same boundary and they assign the same colors to the points outside this
boundary.
6
The main aim of this section is to prove that every polygonal coloring con-
tains every nonequilateral triangle, and to characterize the polygonal colorings
that avoid an equilateral triangle. To achieve this, we need the following defi-
nition:
Definition3.2. Acoloringχ=(B,W)iscalledzebra-like ifithasthefollowing
form: the boundaryofχisadisjointunionofinfinitely manycontinuouscurves
L ;i Z with the following properties (see Fig. 4):
i
∈
(a) There is a unit vector ~x such that for every i Z, L +~x = L . In other
i i
words, the L are invariant upon a translation∈of length 1.
i
(b) For every i Z, the curve Li+1 is a translated copy of Li. Moreover,there
∈
is a unit vector ~y orthogonalto ~x, so that
1 √3
Li+1 =Li+ ~x+ ~y.
2 2
In other words, for an arbitrary boundary point X L , the points Y =
i
∈
X +~x and Z = X + 1~x+ √3~y belong to the boundary as well. Note that
2 2
XYZ is a unit triangle, and that Y Li and Z Li+1.
∈ ∈
(c) For every i Z, the interior of the region delimited by Li Li+1 is colored
with adiffer∈entcolorthanthe interiorofthe regiondelimit∪ed byLi 1 Li.
− ∪
(d) For two points A and B, let θ denote the size of the acute angle formed
AB
by the segment AB and the vector ~x. For every i Z and every two
points A Li and B Li+1, the following holds: AB∈ > 1 if and only if
θ < π.∈ ∈ k k
AB 3
This last condition can also be stated in the following equivalent form: Let
A ∈ Li be an arbitrary point on the boundary. Let B1 = A− 21~x+ √23~y
and B2 = A+ 21~x+ √23~y (the two points B1,B2 belong to Li+1 by the
previous conditions), and let A′ = A+√3~y (so that A′ Li+2). Under
these assumptions, the portion of Li+1 between B1 and∈B2 is contained
inside of the closed lens-shaped region D(A) D(A) and no other point of
′
Li+1 is inside this region. ∩
We stress that a zebra-like coloring is not necessarily polygonal.
3.1 The result
The following theorem is the main result of this section:
Theorem 3.3. For a polygonal coloring χ, the following conditions are equiv-
alent:
(C1) The coloring χ is a zebra-like polygonal coloring.
(C2) The coloring χ has a twin χ which avoids the unit triangle.
′
7
Li+1 B1 B B2
π/3
L ~x
i
A
~y
1~x+√3~y
2 2
Li 1 ~x
−
Figure 4: The boundary of a zebra-like coloring
(C3) For every monochromatic unit triangle ABC, at least one of the three
points A,B and C belongs to the boundary of χ.
Clearly,thecondition(C2)ofTheorem3.3impliesthecondition(C3),sowe
only need to prove that (C1) implies (C2) and that (C3) implies (C1).
The proofis organizedas follows: wefirstprovethat(C3) (C1). This part
⇒
of the proof proceeds in several steps: first of all, we use the condition (C3) to
describe the set ∆(χ) C(A), where A is a boundary point. Then we apply a
∩
continuity argument to extend this information into a global description of χ.
Next, in Theorem 3.19, we prove that every (not necessarily polygonal)
zebra-like coloring has a twin that avoids the unit triangle, which shows that
(C1) (C2), completing the proof of Theorem 3.3.
⇒
Inthelastpartofthis section,weshowthatTheorem3.3implies thatevery
polygonalcoloringcontains a monochromaticcopy T of a given non-equilateral
triangle, with the vertices of T avoiding the boundary.
3.2 The proof
We begin with an auxiliary lemma:
Lemma 3.4. Let q1,q2,q3 be (not necessarily distinct) lines in the plane, not
all three parallel. Then exactly one of the following possibilities holds:
1. The lines q1,q2,q3 intersect at a common point and every two of them
form an angle π.
3
2. ThereexistonlyfinitelymanyunittrianglesABC suchthatA q1,B q2
∈ ∈
and C q3.
∈
Proof. It can be easily checked that the two conditions cannot hold simultane-
ously: in fact, if the three lines satisfy the first condition, then for every point
A q1 whose distance from the other two lines is at most 1 there are points
∈
B q2 and C q3 such that ABC is a unit triangle. We now show that at
∈ ∈
least one of the two conditions holds.
Since the three lines are not all parallel, we may assume that neither q1 nor
q2 is parallel to q3. Consider a Cartesian coordinate system whose y-axis is
8
q3. There exist real numbers a1,a2,b1,b2 such that for i 1,2 we have qi =
(x,y) R2;y =aix+bi . Let ABC be a unit triangle w∈ith{ A=} (x1,y1) q1,
{ ∈ } ∈
B = (x2,y2) q2 and C q3, and assume that A,B,C are in the counter-
clockwise orde∈r (the other∈case is symmetric). Then C = (x1+x2,y1+y2) +
2 2
√23(y1−y2,x2−x1). ThepointC liesonq3,whichimpliesthefollowingequality:
x1+x2 √3(y1 y2)
+ − =0 (1)
2 2
Points A and B are at the distance 1, from which we get
(x1 x2)2+(y1 y2)2 =1 (2)
− −
By combining (1) and (2) and eliminating y1,y2 we get
2
x1+x2 = 3 1 (x1 x2)2 ,
(cid:18) 2 (cid:19) 4 − −
(cid:0) (cid:1)
which yields
3
x21+x22−x1x2 = 4. (3)
Substituting y1 =a1x1+b1 and y2 =a2x2+b2 into (1) gives
1+√3a1 1 √3a2 √3
x1+ − x2+ (b1 b2)=0 (4)
2 2 2 −
If both 1+√3a1 and 1 √3a2 are equal to zero, then the equality (4) degenerates
2 −2
and we get that a1 = 1 , a2 = 1 and b1 = b2, so the first case of the
−√3 √3
statement holds.
In the other case, suppose (wlog) that 1+√3a1 =0. From (4) we can obtain
2 6
that x1 = cx2 + d for some reals c,d. By substituting it into (3) we get a
quadratic equation for the variable x2, where the leading coefficient is equal to
c2−c+1=(c− 12)2+ 34 >0, so there exist at most two possible values for x2,
thusatmosttwopossiblelocationsofB andatmostfourpossibleunittriangles
ABC.
Throughout the rest of this section, we assume that χ is a fixed polygonal
coloringsatisfyingthecondition(C3)ofTheorem3.3. Everyboundarysegment
can be regarded as a common edge of two (possibly unbounded) polygonal re-
gions,oneofwhichiswhiteandtheotherblack. Wechooseanorientationofthe
boundary segments in the following way: a boundary segment with endpoints
A and B is directed from A to B if the white region adjacent to this segment
is on the left hand side from the point of view of an observer walking from A
to B.
Definition 3.5. A boundary point A ∆ is called feasible, if A is not a
boundaryvertex,andtheunitcircleC(A)∈doesnotcontainanyboundaryvertex.
An infeasible point is a point on the boundary that is not feasible.
9
We may easily see that every bounded subset of the plane contains only
finitely many infeasible points.
The first step in the proof of the main result is the description of the set of
all the boundary points at the unit distance from a given feasible point A.
Let A be a fixed feasible point, let s be the boundary segment containing
A. The set ∆ C(A) is finite, by the definition of polygonal coloring; on the
∩
other hand, this set is nonempty, otherwise we could find two points B,C of
C(A) such that ABC is a unit triangle, with B and C in the interior of the
same color class. By shifting the triangle ABC slightly in a suitable direction,
we would obtain a monochromatic unit triangle avoiding the boundary, which
is forbidden by the condition (C3).
Inthefollowingarguments,wewilluseaCartesiancoordinatesystemwhose
originisthepointA,andwhosex-axisisparalleltosandhasthesameorienta-
tion. Weshallassumethatthe x-axisandthe segmentsisdirectedleft-to-right
and the y-axis is directed bottom-to-top. Assuming this coordinate system, we
let P(α,A) denote the point of C(A) with coordinates (cos(α),sin(α)). If no
ambiguity arises, we write P(α) instead of P(α,A).
Lemma 3.6. Let B = P(α) be an arbitrary element of ∆ C(A), let t be the
∩
boundary segment containing B (the segment t is determined uniquely, because
A is a feasible point). Then the segments s and t are parallel.
Proof. For contradiction, assume that s and t are not parallel, let σ (0,π) be
∈
the angular slope of t with respect to the coordinate system established above,
i.e., σ is the angle formed by the lines containing s and t.
First of all, note that the point C = P(α + π) lies on the boundary ∆;
3
otherwise, a sufficiently small translation of the unit triangle ABC in a suit-
able directionwould yield a counterexample to condition (C3) (here we use the
assumption that s and t are not parallel). Let u be the boundary segment
containing C, and let τ be the angular slope of u.
Secondly,wemaydeduce that σ,τ = π,2π ,andthethree linescontain-
{ } {3 3 }
ings,tanduallmeetatonepoint. Ifthiswerenotthecase,thenbyLemma3.4
there would be only finitely many unit triangles with vertices belonging to the
three segments s, t and u. Thus, we could find a unit triangle AB C with
′ ′ ′
A s, B t and C ∆, which is impossible, by the argument presented in
′ ′ ′
the∈previou∈sparagraph6∈. Byrepeatingthisargumentwith α+iπ; i=1,...,5
{ 3 }
in place of α, we obtain the following conclusions:
The six points P(α+ iπ); i=1,...,6 all belong to the boundary ∆.
• { 3 }
The lines passing through the boundary segments containing these six
•
points all meet at one point.
The boundary segments containing P(α), P(α+ 2π) and P(α+ 4π) all
• 3 3
have the same slope.
This is a contradiction, because three parallel segments intersecting a circle in
threedistinctpointscannotbelongtoasingleline,andtwoparallellinesdonot
intersect.
10