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One recent development is the discovery of "zebra-like" colorings that avoid an equilateral triangle; before, it was unknown whether the only 2-coloring to avoid an equilateral triangle is the obvious alternating strip coloring. The relevant paper is http://arxiv.org/abs/math.CO/0701940.

I have written on the subject. In http://www.math.ucsd.edu/~etressle/monotriangle.pdf I show a 3-coloring that avoids the 1,1,2 triangle. I will think about this for 2 colors. I believe it should be impossible, but I will update this later if I have more thoughts. It is easy to see (due to Soifer) that either a 1,1,1 or a 2,2,2 triangle imply that a 1,1,2 triangle exists (for 2 colors, again). It is also known that no two-coloring simultaneously avoids equilateral triangles of sides 1, 2, and 3.

The general conjecture is still wide open, as is the conjecture that 3 colors suffice to avoid any given triangle.

Edit (1 August): I have spent some time trying to prove that no 2-coloring avoids the 1,1,2 triangle, and cannot prove it. If anyone makes progress on this, or knows the answer, or even just wants to work on it with me, please let me know.

One recent development is the discovery of "zebra-like" colorings that avoid an equilateral triangle; before, it was unknown whether the only 2-coloring to avoid an equilateral triangle is the obvious alternating strip coloring. The relevant paper is http://arxiv.org/abs/math.CO/0701940.

In http://nucularpower.com/papers/monotriangle.pdf I show a 3-coloring that avoids the 1,1,2 triangle; I have been unable to find a 2-coloring that does so. It is easy to see (due to Soifer) that either a 1,1,1 or a 2,2,2 triangle imply that a 1,1,2 triangle exists (for 2 colors, again). It is also known that no two-coloring simultaneously avoids equilateral triangles of sides 1, 2, and 3.

The general conjecture is still wide open, as is the conjecture that 3 colors suffice to avoid any given triangle.

Edit, December 2016: I have updated the link to my note. In case the link becomes outdated again, the 3-coloring that avoids a degenerate triangle is just the coloring shown below, where the upper hexagon illustrates how boundaries are colored.

One recent development is the discovery of "zebra-like" colorings that avoid an equilateral triangle; before, it was unknown whether the only 2-coloring to avoid an equilateral triangle is the obvious alternating strip coloring. The relevant paper is http://arxiv.org/abs/math.CO/0701940.

I have written on the subject. In http://www.math.ucsd.edu/~etressle/monotriangle.pdf I show a 3-coloring that avoids the 1,1,2 triangle. I will think about this for 2 colors. I believe it should be impossible, but I will update this later if I have more thoughts. It is easy to see (due to Soifer) that either a 1,1,1 or a 2,2,2 triangle imply that a 1,1,2 triangle exists (for 2 colors, again). It is also known that no two-coloring simultaneously avoids equilateral triangles of sides 1, 2, and 3.

The general conjecture is still wide open, as is the conjecture that 3 colors suffice to avoid any given triangle.

One recent development is the discovery of "zebra-like" colorings that avoid an equilateral triangle; before, it was unknown whether the only 2-coloring to avoid an equilateral triangle is the obvious alternating strip coloring. The relevant paper is http://arxiv.org/abs/math.CO/0701940.

I have written on the subject. In http://www.math.ucsd.edu/~etressle/monotriangle.pdf I show a 3-coloring that avoids the 1,1,2 triangle. I will think about this for 2 colors. I believe it should be impossible, but I will update this later if I have more thoughts. It is easy to see (due to Soifer) that either a 1,1,1 or a 2,2,2 triangle imply that a 1,1,2 triangle exists (for 2 colors, again). It is also known that no two-coloring simultaneously avoids equilateral triangles of sides 1, 2, and 3.

The general conjecture is still wide open, as is the conjecture that 3 colors suffice to avoid any given triangle.

Edit (1 August): I have spent some time trying to prove that no 2-coloring avoids the 1,1,2 triangle, and cannot prove it. If anyone makes progress on this, or knows the answer, or even just wants to work on it with me, please let me know.

One recent development is the discovery of "zebra-like" colorings that avoid an equilateral triangle; before, it was unknown whether the only 2-coloring to avoid an equilateral triangle is the obvious alternating strip coloring. The relevant paper is http://arxiv.org/abs/math.CO/0701940.

I have written on the subject. In http://www.math.ucsd.edu/~etressle/monotriangle.pdf I show a 3-coloring that avoids the 1,1,2 triangle. I will think about this for 2 colors. I believe it should be impossible, but I will update this later if I have more thoughts.

The general conjecture is still wide open, as is the conjecture that 3 colors suffice to avoid any given triangle.

One recent development is the discovery of "zebra-like" colorings that avoid an equilateral triangle; before, it was unknown whether the only 2-coloring to avoid an equilateral triangle is the obvious alternating strip coloring. The relevant paper is http://arxiv.org/abs/math.CO/0701940.

I have written on the subject. In http://www.math.ucsd.edu/~etressle/monotriangle.pdf I show a 3-coloring that avoids the 1,1,2 triangle. I will think about this for 2 colors. I believe it should be impossible, but I will update this later if I have more thoughts. It is easy to see (due to Soifer) that either a 1,1,1 or a 2,2,2 triangle imply that a 1,1,2 triangle exists (for 2 colors, again). It is also known that no two-coloring simultaneously avoids equilateral triangles of sides 1, 2, and 3.

The general conjecture is still wide open, as is the conjecture that 3 colors suffice to avoid any given triangle.