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Colour each point on a circle of radius $\frac{1}{2}$ red or blue, such that the region of blue points has length $1$. Prove that we can inscribe an equilateral triangle in the circle such that all three vertices are red.

I think the Pigeonhole Principle will be involved, but don't quite see how to apply it. The length condition also seems a bit hard to work with, so any hints or suggestions would be much appreciated.

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    $\begingroup$ What does "region of blue points has length 1" mean? Is one of the assumptions that the blue set is measurable? $\endgroup$ – John Hughes Dec 9 '18 at 21:41
  • $\begingroup$ Yes, we are essentially assuming that you can get all the blue points together in a line and measure its length. $\endgroup$ – Prasiortle Dec 9 '18 at 21:53
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Make all the red points that are a distance exactly $\frac {2\pi}3$ away from a blue point blue. The measure of the blue points is now no more than $3$, but the circumference of the circle is $\pi$. There is at least $\pi-3$ of the circle still colored red and any of the red points is on an all red equilateral triangle.

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