If we color the plain with three different colours, then there will always be an equilateral triangle which has three vertices of the same colour.

I have proved it for two colours but I just can't solve this one.

  • $\begingroup$ I'm fairly certain that it has been solved...at least my teacher says he knows a proof $\endgroup$ – user327929 Apr 26 '16 at 17:10

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This needs van der Waerden's Theorem. There exists $N(r,k)$ such that in any line of $N$ equally spaced points which are each coloured with one of $r$ colours, we can always find $k$ equally spaced points which are monochromatic.

Let $N=N(2,3)$ and $M=N(3,N+1)$. Then if we take any $M$ equally spaced collinear points in the 3-coloured plane, we can find $N$ equally spaced monochromatic points along a line segment $L$. Suppose they are coloured 1.

That gives us a large triangle of equally spaced points in the equilateral triangle with side $L$. If any of them is also coloured 1, then we have a monchromatic equilateral triangle with two points in $L$. So assume they are all coloured 2 or 3.

The next line up in the triangle has $N$ points, so we can find 3 of them which are equally spaced and monochromatic. Suppose they are coloured 2. Call them $A,B,C$. Further up in the large triangle we have $D$ so that $ABD$ is equilateral. Now if it is not monochromatic $D$ must be coloured 3. Similarly, the point $E$, with $BCE$ equilateral must also be coloured 3. But now the point $F$ further up in the large triangle makes $DEF$ equilateral.

But however it is coloured, $F$ must be in a monochromatic equilateral triangle. The other two vertices of $DEF$ are coloured 2, the other two vertices of $BDF$ are coloured 3, and there are two vertices in $L$ forming an equilateral triangle with $F$ which are coloured 1.

  • $\begingroup$ Sorry...I posted the question incorrectly...I have edited it, and your solution is incorrect for the edited problem... $\endgroup$ – user327929 Apr 26 '16 at 16:32
  • $\begingroup$ All vertices of the triangle are the same colour $\endgroup$ – user327929 Apr 26 '16 at 16:35
  • $\begingroup$ In your solution you use that the center point must be 3 but that isn't true... $\endgroup$ – user327929 Apr 26 '16 at 16:36
  • $\begingroup$ @BBB Sorry for the delay; I have finally fixed this proof. $\endgroup$ – almagest Apr 26 '16 at 18:05

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