A Euclidean Ramsey result in the plane

An old question in Euclidean Ramsey theory asks, if the points in the plane are red-blue coloured, does there always exist a red pair of points at unit distance or five blue points in line separated by unit distances? An elementary proof answers this question in affirmative.


Introduction
Many problems in Euclidean Ramsey theory ask, for some d ∈ Z + , if E d is coloured with r ≥ 2 colours, does there exist a colour class containing some desired geometric structure? Research in Euclidean Ramsey theory was surveyed in [2][3][4] by Erdős, Graham, Montgomery, Rothschild, Spencer, and Straus; for a more recent survey, see Graham [5]. Say that two geometric configurations are congruent iff there exists an isometry (distance preserving bijection) between them. For d ∈ Z + , and geometric configurations F 1 , F 2 , let the notation E d → (F 1 , F 2 ) mean that for any red-blue coloring of E d , either the red points contain a congruent copy of F 1 , or the blue points contain a congruent copy of F 2 . For a positive integer i, denote by ℓ i the configuration of i collinear points with distance 1 between consecutive points. One of the results in [3] states that E 2 → (ℓ 2 , ℓ 4 ). (1) In the same paper, it was asked if E 2 → (ℓ 2 , ℓ 5 ), or perhaps a weaker result holds: The result (1) was generalised by Juhász [7], who proved that if T 4 is any configuration of 4 points, then E 2 → (ℓ 2 , T 4 ). Juhász (personal communication, 10 February 2017) informed the author that Iván's thesis [6] contains a proof that for any configuration T 5 of 5 points, E 3 → (ℓ 2 , T 5 ) (which implies that E 3 → (ℓ 2 , ℓ 5 )). Arman and Tsaturian [1] proved that E 3 → (ℓ 2 , ℓ 6 ). In this paper, it is proved that E 2 → (ℓ 2 , ℓ 5 ): Theorem 1.1. Let the Euclidean space E 2 be coloured in red and blue so that there are no two red points distance 1 apart. Then there exist five blue points that form an ℓ 5 .

Proof of Theorem 1.1
The proof is by contradiction; it is assumed that there are no five blue points forming an ℓ 5 . The following lemmas are needed.
Lemma 2.1. Let E 2 be coloured in red and blue so that there is no red ℓ 2 . If there is no blue ℓ 5 , then there are no three blue points forming an equilateral triangle with side length 3 and with a red centre.
Proof. Suppose that E 2 is coloured in red and blue so that there is no red ℓ 2 and no blue ℓ 5 . Suppose that blue points A, B and C form an equilateral triangle with side length 3 and with red centre O. Consider the part of the unit triangular lattice shown in Figure 1(a). The points D, E, F , G are blue, since they are distance 1 apart from O. The point X is red; otherwise XADEB is a red ℓ 5 . Similarly, Y is red (to prevent red Y AF GC). Then X and Y are two red points distance 1 apart, which contradicts the assumption. Figure 1: Red points are denoted by diamonds, blue points are denoted by discs.
Lemma 2.2. Let E 2 be coloured in red and blue so that there is no red ℓ 2 . If there is no blue ℓ 5 , then there are no three red points forming an equilateral triangle with side length √ 3 and with a red centre.
Proof. Suppose that E 2 is coloured in red and blue so that there is no red ℓ 2 and no blue ℓ 5 . Suppose that blue points A, B and C form an equilateral triangle with side length 3 and with red centre O. Let A ′ , B ′ , C ′ be the images of A, B and C, respectively, under a rotation about O so that AA ′ = BB ′ = CC ′ = 1 (see Figure 1(b)). Then A ′ , B ′ , C ′ are blue and form an equilateral triangle with side length √ 3 and red center O, which contradicts the result of Lemma 2.1.
Define T 3 , T 4 , T 5 , T 6 , T 7 to be the configurations of three, four, five, six and seven points (respectively), depicted in Figure 2 (all the smallest distances between the points are equal to √ 3).  Proof. Suppose that E 2 is coloured in red and blue so that there is no red ℓ 2 and no blue ℓ 5 . Suppose that A, B, C, D, E, F and G are red points forming a T 7 (as in Figure 3). Let X be the reflection of F in BC. Let X ′ , A ′ , F ′ be the images of X, A, F , respectively, under the clockwise rotation about B such that XX ′ = AA ′ = F F ′ = 1. Since A and F are red, A ′ and F ′ are blue. If X ′ is blue, then X ′ A ′ F ′ is a blue equilateral triangle with side length 3 and red center B, which contradicts the result of Lemma 2.1. Therefore, X ′ is red. Let X ′′ , D ′′ , F ′′ be the images of X, D, F , respectively, under the clockwise rotation about C such that XX ′′ = DD ′′ = F F ′′ = 1. Since D and F are red, D ′′ and F ′′ are blue. If X ′′ is blue, then X ′′ D ′′ F ′′ is a blue equilateral triangle with side length 3 and red center C, which contradicts the result of Lemma 2.1. Therefore, X ′′ is red. Since X ′ can be obtained from X ′′ by the clockwise rotation through 60 • about X, XX ′ X ′′ is a unit equilateral triangle, hence X ′ X ′′ is a red ℓ 2 , which contradicts the assumption of the lemma.  Proof. Suppose that E 2 is coloured in red and blue so that there is no red ℓ 2 and no blue ℓ 5 . Let A, B, C be three red points forming a T 3 . Consider the unit triangular lattice depicted in Figure 4. Suppose that there is no red point D such that A, B, C, D form a T 4 . Then points X, Y , Z are blue. Points E, F , G, H, I, J are blue, since each of them is distance 1 apart from a red point. If the point K is red, then the points L and M are blue and LM Y GH is a blue ℓ 5 . Therefore, K is blue. Then N is red (otherwise KJIZN is a blue ℓ 5 ), hence P and Q are blue, which leads to a blue ℓ 5 P QF EX. A contradiction is obtained, therefore there exists a red point D such that A, B, C, D form a T 4 . Let A, B, C, D form a red T 4 . Consider the part of the unit triangular lattice depicted in Figure 5. Suppose that there is no red point E such that A, B, C, D, E form a T 5 . Then the points X, F and G are blue. Points H, I, K, L, M , N are blue, since each of them is distance 1 apart from a red point. Point P is red (otherwise F HIGP is a blue ℓ 5 ), therefore Q and R are blue. Then X, N , M , Q, R form a blue ℓ 5 , which gives a contradiction. Hence, there exists a red point E such that A, B, C, D, E form a T 5 .  Figure 5 Let A, B, C, D, E form a T 5 ( Figure 6). Suppose that F is blue. By Lemma 2.2, points X and Y are blue (otherwise X, E, C (Y , A, D) form a red triangle with side length 3 and red center B). Points G, H, I, J, K, L, M , N are blue, since each one of them is at distance 1 from a red point. If point P is blue, then Q is red (otherwise QP KLF is a blue ℓ 5 ), U and T are blue and form a blue ℓ 5 with points G, H and X. Therefore, P is red. Similarly, R is red (otherwise S is red and V W JIY is a blue ℓ 5 ). Then A, B, C, D, E, P and R form a red T 7 , which is not possible by Lemma 2.3. Therefore, F is red and A, B, C, D, E, F form a red T 6 . Lemma 2.5. Let E 2 be coloured in red and blue so that there is no red ℓ 2 . Let L be a unit triangular lattice that contains three red points forming a T 3 . If there is no blue ℓ 5 , then the colouring of L is unique (up to translation or rotation by a multiple of 60 • ), and is depicted in Figure 7.

Figure 7
Proof. Suppose that E 2 is coloured in red and blue so that there is no red ℓ 2 and no blue ℓ 5 . Suppose there exist three red points of L that form a T 3 . By Lemma 2.4, it may be assumed that there is a red T 6 . Denote its points by A, B, C, D, E, F (see Figure 8). It will be proved that the translate A ′ B ′ C ′ D ′ E ′ F ′ of ABCDEF by the vector of length 5 collinear to − − → AD is red. Consider the points shown in Figure 8. Since A, D and F are red, by Lemma 2.2, I is blue. Since C, F and D are red, by Lemma 2.2, J is blue. Points K, L, M , N are blue, since each one is distance 1 apart from a red point. If R is red, then both P and Q are blue and form a blue ℓ 5 with K, L and I. Therefore R is blue. Then the point A ′ is red (otherwise A ′ JN M R is a red ℓ 5 ). Since S 1 , S 2 , S 3 , S 4 are blue (as distance 1 apart from red points D and A ′ ), B ′ is red. Similarly, F ′ is red. Points V and W are blue as they are distance 1 apart from C. Points U is blue by Lemma 2.2 (since A, D and B are red). If X is red, then X 1 and X 2 are blue and a blue ℓ 5 U V W X 1 X 2 is formed. Therefore, X is blue. Similarly, Y is blue. By Lemma 2.4, A ′ B ′ F ′ must be contained in a red T 6 , and since X and Y are blue, the only possible such Similarly, the translates of ABCDEF by vectors of length 5 collinear to − − → EB and − − → CF are red. By repeatedly applying the same argument to the new red translates, it can be seen that all the translates of ABCDEF by a multiple of 5 in L are red. All the other points are blue, as each one is distance 1 apart from a red point. Hence, the colouring as in Figure 7 is obtained.
Lemma 2.6. Let E 2 be coloured in red and blue so that there is no red ℓ 2 . Let L be a unit triangular lattice that does not contain three red points forming a T 3 . If there is no blue ℓ 5 , then the colouring of L is unique (up to translation or rotation by a multiple of 60 • ), and is depicted in Figure 9.

Figure 9
Proof. Suppose that E 2 is coloured in red and blue so that there is no red ℓ 2 and no blue ℓ 5 . If L does not contain a red point, then any ℓ 5 is blue, therefore L contains a red point A. By Lemma 2.1, one of the points of L at distance √ 3 to A is red (otherwise the three such points form a blue triangle with side length 3 and red centre A). Denote this point by B (Figure 10). Since L does not contain a red T 3 , the points D and G are blue. Points E, F , I, H, K, J are blue, since they are distance 1 apart from B. Then the point B ′ is red (otherwise blue ℓ 5 DEF GB ′ is formed). Point N is 1 apart from B ′ , hence blue. Then C and A ′ are red (otherwise a blue ℓ 5 is formed). By repeating the same argument for points B and C, B and A (instead of A and B), and so on, it can be shown that any node of L on the line AB is red. Similarly, since A ′ and B ′ are both red, any node of L on the line A ′ B ′ is red. By the same argument, A ′′ , B ′′ and any node on the line containing them is red; A ′′′ , B ′′′ and any node on the line containing them is red, and so on. By colouring all point distance 1 apart form red points blue, the colouring in Figure 9 is obtained. Proof of Theorem 1.1. Let the Euclidean space E 2 be coloured in red and blue so that there are no two red points distance 1 apart. Suppose that there are no five blue points that form an ℓ 5 . Then there is a red point A. Consider two points B and C, both distance 5 apart from A, such that |BC| = 1. At least one of the points B and C (say, B) is blue. Consider the unit triangular lattice L that contains A and B. By Lemma 2.5 and Lemma 2.6, L is coloured either as in Figure 7 or as in Figure 9. But neither one of the colourings contains two points of different colour distance 5 apart, which gives a contradiction. Therefore, there exist five blue points that form an ℓ 5 .

Acknowledgements
The author would like to thank Ron Graham and Rozália Juhász for providing information about the current state of the problem, and David Gunderson for valuable comments.