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In a similar vein to a question I asked a few days ago:

Do all two-colorings of $\mathbb{R}^2$ contain three points of the same color which form the vertices of a degenerate triangle of side-lengths (1,1,2)?

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You might want to reread your question and add a negation somewhere. – Phira May 18 '12 at 7:10
Oops - swapped an Exists for a For All. Thanks @Phira. – Ternary May 18 '12 at 20:49
So a monochromatic (1,1,1) implies a (1,1,2) fairly quickly, as does a (2,2,2). But our abundance of $\sqrt{3}$ length equilaterals gets us nothing. – Ternary May 25 '12 at 18:42

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