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If every point on a circle of radius $R$ in $\mathbb{R}^{2}$ were colored one of two colors, is there necessarily two points that are of the same color and of distance $R$ apart? what about $>2$ colors?

On the $2$-color case, I'm thinking that if we have two points of different colors on the circle, say red and blue, then consider the chord of length $R$ from the blue point to another point on the circle. This new point must then be red. Continuing, we get a hexagon if I'm not mistaken, which doesn't help. But we can place another hexagon inscribed in the circle with the same colorings, ad infinitum. Intuitively, this leaves me to believe that there must be two points of the same color of distance $R$ apart on the circle due to some induction or something. I'm not sure! Does this hold for more than two colors?

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Given the point $(\cos \phi,\sin\phi)$, colour it red if $\phi\in[0,\frac\pi6[\cup [\frac\pi 3,\frac\pi2[\cup [\frac{2\pi} 3,\frac{5\pi}6[$ and blue otherwise. This is just your hexagon repeated over an interval.

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The answer to the question in the title is No. The circle is the disjoint union of six-element sets consisting of regular hexagon vertices. You can color the vertices of any such hexagon alternatively red and blue. Any two vertices of different hexagons cannot have distance $R$; therefore the chosen colorings do not interfer.

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