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Shader 1976 all right triangles are ramsey e2

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corollary_4: Shader's corollary that every triangle with sides a, b and (b^2 + 2a^2)^{1/2} with 2b > a is Ramsey: every two-coloring of the plane contains a monochromatic triangle congruent to it.

corollary_5: Shader's corollary that every triangle with sides a, b and (4b^2 - a^2)^{1/2} with (3/2)^{1/2} b < a < (5/2)^{1/2} b is Ramsey: every two-coloring of the plane contains a monochromatic triangle congruent to it.

lemma_1: Shader's lemma that for any real number a and any two-coloring of the plane there is a monochromatic equilateral triangle of side ka for some k in {1, 3, 5, 7}, where k may depend on a.

theorem_2: Shader's theorem that every right triangle is Ramsey in the plane: every two-coloring of the plane contains a monochromatic triangle congruent to it.

theorem_3: Shader's theorem that for every parallelogram P and every two-coloring of the plane there is a parallelogram congruent to P with three vertices of one color.


Leslie E. Shader, All Right Triangles Are Ramsey in E^2!. Journal of Combinatorial Theory, Series A 20 (1976), 385-389. doi:10.1016/0097-3165(76)90036-4. The file prints "Copyright © 1976 by Academic Press, Inc. All rights of reproduction in any form reserved." in its first-page footer, every other right reserved.

A triangle is called Ramsey in the plane if every two-coloring of the plane contains a monochromatic congruent copy. The paper proves four results (p. 385, labelled on pp. 388-389): every right triangle is Ramsey (Theorem 2); every parallelogram has a congruent copy in which three of the four vertices share a color (Theorem 3); each triangle with sides (a, b, sqrt(b^2 + 2a^2)) and 2b > a is Ramsey (Corollary 4); and each triangle with sides (a, b, sqrt(4b^2 - a^2)) and sqrt(3/2) b < a < sqrt(5/2) b is Ramsey (Corollary 5). The engine is Lemma 1 (p. 385): given a real a and a two-coloring of the plane, some equilateral triangle with side ka, for one of k = 1, 3, 5, 7 (k may depend on a), is monochromatic. By [2, Theorem 1] (Erdos) it suffices to find monochromatic copies of two triangles with odd integer sides (3, 5, 7 and 7, 15, 13) or of their odd multiples, and a case analysis on the colors of the points with integer coordinates in the frame spanned by an equilateral triangle of side 8 (Fig. 1) supplies them. Theorem 2 then follows by the "ladder" technique of [2], and the corollaries by applying Theorem 3 to a parallelogram and a rhombus. For problem 173 this is the statement-cited primary source for the right-triangle case, but it is only a special-case result and does not establish the conjecture, repeated here from earlier work [1] (Erdos, Graham, Montgomery, Rothschild, Spencer and Straus), that every non-equilateral triangle is Ramsey.

Source: https://doi.org/10.1016/0097-3165(76)90036-4.

Bears on. #173: Theorem 2 and Corollaries 4 and 5 show that no right triangle, no triangle with sides aa, bb, b2+2a2\sqrt{b^2+2a^2} and 2b>a2b>a, and no triangle with sides aa, bb, 4b2−a2\sqrt{4b^2-a^2} and 3/2 b<a<5/2 b\sqrt{3/2}\,b<a<\sqrt{5/2}\,b can be the exceptional triangle of a two-coloring of the plane, and Lemma 1 that no two-coloring misses the equilateral triangles of all four sides aa, 3a3a, 5a5a, 7a7a. The paper says nothing about whether one coloring can miss two other triangles, which is the question.

Results. The printed statement of Corollary 5 gives the third side as 4b2−a24b^2-a^2, without the square root, while p. 385 and the proof give (4b2−a2)1/2(4b^2-a^2)^{1/2}; on p. 385 the lower end of its range is printed without the factor bb. The result pages record both.

  • Lemma 1 (p. 385): monochromatic equilateral triangle of side kaka, k∈{1,3,5,7}k\in\{1,3,5,7\}.
  • Theorem 2 (p. 388): all right triangles are Ramsey.
  • Theorem 3 (p. 388): every parallelogram has a congruent copy with three vertices of one color.
  • Corollary 4 (p. 389): the triangles (a,b,(b2+2a2)1/2)(a,b,(b^2+2a^2)^{1/2}), 2b>a2b>a, are Ramsey.
  • Corollary 5 (p. 389): the triangles (a,b,(4b2−a2)1/2)(a,b,(4b^2-a^2)^{1/2}), (3/2)1/2 b<a<(5/2)1/2 b(3/2)^{1/2}\,b<a<(5/2)^{1/2}\,b, are Ramsey.

Read status: claims checked for Lemma 1, Theorems 2 and 3 and Corollaries 4 and 5, read clause by clause on the page images of the print; the proofs read for structure only. The reduction and ladder technique of reference [2] are cited, not proved, in the paper and were not read. A second reader checked the result pages' statements, hypotheses, labels and pages against the print; the proofs were not independently reviewed.

No file of this source is held: no license on record permits its redistribution, and the card cites the edition it names above.