Prove there are 3 points on the circle having same colour

Hint:

Take regular pentagon. The three points are the same color.

In the wonderfully clever solution given, starting with that pentagon, the two possible monochromatic triangles are not congruent. It may be interesting to note that this is of necessity.

Say an $\alpha$-triangle is an isoceles triangle in which the two sides of equal length meet at an angle $\alpha$.

Fix an angle $\alpha$. There exists a coloring of the points of the cirle with two colors, such that no $\alpha$-triangle is monochromatic.

Say the circle is the unit circle $S$ in the complex plane. Given $\alpha$, there exists a complex number $\beta$ with $|\beta|=1$ such that the vertices of any $\alpha$-triangle are $z,\beta z,\beta^2z$ for some $z$. For $z\in S$ define $$C_z=\{\beta^nz:n\in\Bbb Z\}.$$Then the $C_z$ form a partition of $S$; choose a coloring so that $\beta^nz$ and $\beta^{n+1}z$ always have different colors.

Oops Unless the sequence $\beta^n z$ is periodic with period $k$, for $k$ odd. In that case color $z,\dots,\beta^{k-1}z$ with alternating colors. (So now $\beta^{k-1}z$ and $\beta^{k}z$ have the same color, but it doesn't matter, there are no three consecutive $n$ for which $\beta^nz$ all have the same color.) Thanks to stewbasic for noticing the gap.

Not anywhere near as elegant as the five-equidistant-point proof, but I've a nine-point brute-force proof.

Premise: any point on a circle will form an icoseles triangle with points placed equidistantly either side.

Pick any number of points separated by distance X.

If any three sequential points are red (111) or blue (000) then there's an isosceles triangle between those three points, of length 2X. So we need a sequence with no more than two of each color, ever.

But by the same logic, any sequence that repeats with period P, will have an isoceles triangle by the beginning of the third repeat, of length 2PX.

Which means no short sequence can possibly be made which, when repeated, prevents the triangle.

So if we find the longest binary number made of alternating (1 or 11) and (0 or 00) where no element appears three times separated by the same interval, and we find it's infinitely long, then we've got a winner. This seems unlikely.

We'll start with all 8 possible combinations of "1, (0 or 00), (1 or 11), (0 or 00), 1", and add more to each until they repeat, or it becomes obvious that they will not.

1 10101 - repeat, 1x1x1.

2 101001 -
    10100101 -
      101001010 - repeat, xxxx0x0x0.
      101001011 - repeat, xx1xx1xx1.
    10100100 - repeat, x0xx0xx0.
    10100110 - repeat, x0xx0xx0.

3 1011001 -> repeat, 1xx1xx1

4 101101 - 
    10110101 - repeat, xxx1x1x1
    10110100 - repeat, x0xx0xx0

5 100101 - mirror of 101001.

6 1001001 - repeat, 1xx1xx1.

7 1001101 - mirror of 1011001: repeat, 1xx1xx1.

8 10011001 -
    100110010 - repeat, xx0xx0xx0
    100110011 - repeat, 1xxx1xxx1

There is no binary number longer than 9 digits in which neither 1 or 0 appears twice at a set interval. Which means that picking nine equidistant points around the circle is sufficient to prove the principle.