Frankl 1986 all triangles are ramsey

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remark_p779: Frankl and Rödl's concluding remark that the symmetric trapezoid with sides sqrt(10), sqrt(8), sqrt(10), sqrt(2) and diagonals sqrt(14) is Ramsey, that the product theorem then gives infinitely many more Ramsey symmetric trapezoids, and that the authors could prove no pentagon Ramsey.

theorem_1: Frankl and Rödl's theorem that every triangle is Ramsey: for each triangle and each number of colors r, every r-coloring of a Euclidean space of high enough dimension has a color class containing a congruent copy of the triangle.

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Peter Frankl, Vojtech Rödl, All triangles are Ramsey. Transactions of the American Mathematical Society 297 (1986), 777-779. doi:10.1090/S0002-9947-1986-0854099-6. The file prints "©1986 American Mathematical Society" on p. 777 and, in every page footer, "License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use", every other right reserved.

Theorem 1 states that every triangle is Ramsey: given a triangle ABC and any r

= 2, for n sufficiently large every r-coloring of R^n contains a monochromatic congruent copy of ABC. The proof runs in three stages. Stage 1 realizes the very obtuse triangles with sides sqrt(2t), sqrt(2t), sqrt(8t-6) by mapping (2t-1)-subsets of {1,...,n} to lattice-like points and applying Ramsey's theorem for l-subsets; Stage 2 upgrades this to all isosceles triangles by rotating ABC about a side and using the product theorem of Erdos, Graham, Montgomery, Rothschild, Spencer and Straus; Stage 3 reaches arbitrary triangles by a projection and continuity argument controlling the ratio tan(alpha)/tan(beta). Concluding remarks note that the method of Stage 1 yields some symmetric trapezoids (sides sqrt(10), sqrt(8), sqrt(10), sqrt(2)), that the authors could not settle any pentagon and that the dimensions produced grow with the configuration; they also announce, without proof here, that all simplices are Ramsey, deferring the less elementary proof to a later paper (p. 779). Read as a scope check for problem 173: the theorem is high-dimensional and holds for every finite color count, so it does not address the two-dimensional two-color statement of #173.

Source: https://www.renyi.hu/~pfrankl/1986-3.pdf.

Read status: claims checked. The definition of a Ramsey set, Theorem 1 and the abstract (p. 777) and the concluding remark on symmetric trapezoids (pp. 778--779) were read clause by clause against the print; the proof of Theorem 1 (pp. 777--778) was read for its structure.

Result pages.

• Theorem 1 (p. 777): every triangle is Ramsey.
• Concluding remark (p. 779): a symmetric trapezoid with sides $\sqrt{10},\sqrt8,\sqrt{10},\sqrt2$ is Ramsey, and the product theorem gives infinitely many more.

Bears on.

• #174: Theorem 1 puts every triangle, and the remark on p. 779 one symmetric trapezoid and a family built from it, in the class of Ramsey sets in the sense of the problem's statement; neither gives a characterization of the Ramsey sets.
• #173: scope only. Theorem 1 lets the dimension grow with the triangle and the number of colors, so it says nothing about two-colorings of the plane.

Results read. The stages are intermediate steps of the proof of Theorem 1 and are summarized on its result page.

• Theorem 1: All triangles are Ramsey: for every triangle and every r >= 2 there is an n_0 such that any r-coloring of R^n, n >= n_0, contains a monochromatic congruent copy.
• Stage 1: The triangle with sides sqrt(2t), sqrt(2t), sqrt(8t-6) is Ramsey for all t >= 2, via a point encoding of (2t-1)-subsets and Ramsey's theorem.
• Stage 1': For integers p, q and any eps > 0 there are Ramsey triangles with angles alpha, beta satisfying |tan alpha / tan beta - p/q| < eps and alpha + beta < eps.
• Stage 2 and 2': All isosceles triangles are Ramsey, and a triangle is Ramsey whenever its orthogonal projection onto a plane through one side is, giving the projection transfer used in Stage 3.
• Concluding remarks: Some symmetric trapezoids (sides sqrt(10), sqrt(8), sqrt(10), sqrt(2), diagonals sqrt(14)) are Ramsey; no pentagon could be handled. That all simplices are Ramsey is announced only, with the proof deferred to a later paper.

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