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Source. Theorem 1, the remark after it and Corollaries 2 and 3, p. 563; the sets and , pp. 563--564; Corollary 4, p. 564; of P. Erdős, R. L. Graham, P. Montgomery, B. L. Rothschild, J. Spencer and E. G. Straus, Euclidean Ramsey Theorems, III, Infinite and Finite Sets (Keszthely 1973), Colloq. Math. Soc. János Bolyai 10, North-Holland (1975), 559--583, as identified on the source card.
Notation
The paper works in the plane with two colors and three-point sets (p. 559). says that every two-coloring of has a monochromatic congruent to ; for a fixed two-coloring , says that some congruent to is monochromatic under (p. 562). and refer to the triangle with sides . says that has an -triangle whose -side has both endpoints of one color and whose third point has the other color (p. 561). A two-coloring is proper when it is not a one-coloring (p. 573).
The paper puts and, for each two-coloring , lets be the set of triples with no monochromatic triangle of sides , , (pp. 563--564).
Statement
Theorem 1 (p. 563). Let be a triangle with sides , , , and let , , be the equilateral triangles of sides , , . Then holds if and only if at least one of , , holds.
The paper calls it a strengthening of Theorem 8 of Part I (Euclidean Ramsey Theorems I). In the notation above it says that exactly when , and all lie in .
Remark (p. 563, credited to R. M. Robinson). The six copies of in the proof are like-oriented, so the proof gives more: if has a monochromatic like-oriented congruent copy under , it also has a monochromatic opposite-oriented one. The paper does not know the analogue for bichromatic copies.
Corollaries (pp. 563--564).
- Corollary 2: if has sides and holds, then holds for every triple with sides where .
- Corollary 3: if fails but holds for a triple with sides , then holds for every triple with sides , .
- Corollary 4: let be an -triangle with , let be a two-coloring of and suppose ; then and . The paper notes after the proof that or can replace .
The paper also observes (p. 564) that by Theorem 1, Conjecture 3 is equivalent to for every .
Proof pointer
P. 563, from Figure 1 (p. 564): eight points carry six triangles with sides and six equilateral triangles, two of each side , , , arranged so that, as in Theorem 8 of Part I, a monochromatic one among the equilateral six forces a monochromatic one among the other six; the converse is the symmetric argument.
Read depth. Claims checked: the statements were read clause by clause on the printed pages; the proof was read for its structure only.
Bears on
- Problem 173: the theorem reduces the problem to equilateral triangles. A two-coloring misses a triangle exactly when it misses the equilateral triangles on all three of its side lengths, so the problem is equivalent to the statement that no two-coloring of the plane misses equilateral triangles of two different sides. The theorem decides no triangle by itself.