Problem. Suppose every point of a circle (with a fixed radius) has been colored either red or blue. Does there exist an equilateral triangle whose 3 vertices are on the circle and share the same color? In more brief terms, does there exist a monochromatic inscribed equilateral triangle inside a $2$-colored circle?
Context. When the conclusion is weakened so that we ask for a monochromatic isosceles triangle instead of an equilateral triangle, the problem is much easier to solve: we can take a regular pentagon inscribed inside the circle. By the pigeonhole principle, three of the five vertices are colored same: those three vertices clearly form an isosceles triangle.
Thank you for your time!
Best Answer
Each point on the circle lies in exactly one equilateral triangle. So the circle is a disjoint union of the inscribed equilateral triangles. Now we choose one point from each triangle to be red, and color the others blue. By construction no triangle is monochromatic.
However, this construction crucially uses the axiom of choice, so you may worry that the resulting coloring is pathological, and a natural follow-up is whether any "nice" coloring (read: a measurable coloring, say) has this property.
It turns out the answer is still no, since two points are in the same inscribed triangle if and only if they're in the same orbit of the $\mathbb{Z}/3$ action on the circle where we rotate by $\frac{2\pi}{3}$. This is a finite (in particular compact, polish) group so its orbit equivalence relation is "smooth". In particular, there's a borel-measurable choice function on the equivalence classes, so that the above construction can actually be done in a borel-measurable way.
Edit: As Michal Adamaszek pointed out in the comments, you can make this extremely concrete.
Say we have a triangle, as shown below:
In the previous section we claimed there's a borel-measurable function choosing one point from each triangle. It's not hard to convince yourself that "take the leftmost point" is a borel thing to do, and is well defined except in the single case shown below where there's a tie:
In this case we can take the upper point.
So then we can color the circle so that the leftmost point of every triangle is red, and the other two points are blue (except for the one triangle with a tie, in which case we color the upper point red). This will give a borel-measurable coloring where, by construction, no triangle is monochromatic!
Here's an example for a single triangle:
And here's the full coloring:
Notice that every triangle has exactly one red and two blue vertices, so that there is no monochromatic triangle (as desired). In hindsight, this is probably quite easy to come up with directly, but this shows some techniques that are used to solve similar problems that may be harder to visualize.
I hope this helps ^_^