I don't know whether this probabilistic method can be accepted as a solution: We prove that the triangle with angles $120,30,30$ and sides $\sqrt 3,1,1$ is good. Firstly note that the center of an equilateral triangle of side $\sqrt 3$ has color opposite to the color that most vertices of the triangle are colored (or else we get our monochromatic triangle).(1) That means if $O$ is an arbitrary point and $ABC$ is an equilateral with side $\sqrt 3$ with center $O$ then at least $2$ of segments $OA,OB,OC$ have endpoints of opposite colors.(2) Now "integrating" that's what we see: Denote by $r$ the probability that from a white point one reaches a black point by a $\sqrt 3$ translation(the point and the translation ar taken arbitrarily). Analogously define $s$ for black points. "Integrating" the previous argument (2) we get $r,s\geq \frac 23$. Now if $t$is the probability that from a given white point one reaches a black one by a translation by magnitude $2\sqrt 3$ is as easily seen $r(1-s)+s(1-r)=r+s-2rs \geq \frac 23+\frac 23-(\frac 23)^2=\frac 89>\frac 56$. Now this means particularly that one can find a regular hexagon $ A B C D E F$ with side $2\sqrt 3$ and center $O$ is white, while at least $5$ of $A,B,C,D,E,F$ are black. WLOG suppose $A,B,C,D$ are black. Then the centers $X,Y$ of $OAB,OCD$ are white by (1). If we take $Z$ the center of $OEF$ then the triangle $XYZ$ with center $O$ contradicts (1) QED????

Probability is not a magic wand. You cannot use it until you define probabilistic space.

Well no wonder I had such a hard time trying to show that no such triangle exists! I think you're right about that triangle, iura, but we don't need such extreme measures (like probability spaces we're not at all sure about ) in order to prove it. If a $\sqrt 3$ equilateral triangle is monochromatic, (say it has color $r$), then its center and the reflections of its center in two sides must have color $b$, and we have a $1,1,\sqrt 3$ monochromatic triangle. This means that if no $1,1,\sqrt 3$ is monochromatic, then no $\sqrt 3,\sqrt 3,\sqrt 3$ is either. Now construct a network of $\sqrt 3$ equilateral triangles (a tesselation of the plane)+their centers. You'll easily find that all pairs of centers distance $2$ apart are monochromatic. Any two points of the plane distance $2$ apart can be seen as two centers in such a network, so pairs of points distance $2$ apart are monochromatic. From here we easily get a contradiction (we find the entire plane to have the same color). I hope it's correct, but if it's not, forgive me: I just woke up and I'm a bit confused . [in the above, "an $a,b,c$" means a triangle with sides $a,b,c$, and "an $\ell$ equilateral triangle" means equilateral triangle with side $\ell$]

grobber wrote: I hope it's correct, but if it's not, forgive me: I just woke up and I'm a bit confused . Just woke up? What was your local time?

I don't know.. About 10 AM. But I went to sleep at about 4 AM .

grobber wrote: I don't know.. About 10 AM. But I went to sleep at about 4 AM . Analogously

Can it be solved in a such way ? First we always can find a segment with vertices of the same color with a given length (just construct equilateral triangle, from three vertices one pair will satisfy condition). Let $A_{1}A_{2}A_{3}A_{4}A_{5}A_{6} $ be a regular hexagon. We choose triangle $A_{1}A_{2}A_{4}$ , then if we find a segment $A_{1}A_{4}$ with two vertices of the same color, assuming that contradiction exists we get $A_{2},A_{3},A_{5},A_{6}$ are of another color so $A_{2}A_{3}A_{5}$ = $A_{1}A_{2}A_{4}$ satisfyies condition. q.e.d.

I can't understand sentence there three "then" are in row without any punctuation marks. Is there some meaning? I don't know.

I think, prowler means the following: without loss of generality, $A_1$ and $A_4$ are red, hence all other vertices are blue and three of them form desired blue triangle. It's true. The other possible solution: some 4 of 7 vertices of regular 7-gon are of the same clour. Then they form a triangle with angles $\pi/7,\,2\pi/7,\,4\pi/7$.