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!

  • $\begingroup$ I have removed the tag "discrete geometry" : it has nothing to do here. $\endgroup$
    – Jean Marie
    Apr 26 at 6:48
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    $\begingroup$ Two different inscribed equilateral triangles have disjoint vertex sets. $\endgroup$ Apr 26 at 6:55
  • $\begingroup$ By "point" are you referring to an arc which is infinitesimally small? And they are colored red or blue alternately? The question isnt very clear. $\endgroup$
    – 44yu5h
    Apr 26 at 6:56
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    $\begingroup$ @44yu5h: I believe the question is already clear. I am happy to elaborate further though. Suppose our circle is $C$, and is defined by the equation $x^2+y^2=1$ in the plane. Now, we color every point on this circle with either blue or red (the coloring is fixed, but we have no control over it). I don't know what you mean by "alternately". Imagine that coloring is given to us by someone. If true, our goal is to show that no matter what the coloring is, there are three points on the circle that have the same color, and when you join these three points, you get an equilateral triangle. $\endgroup$
    – Prism
    Apr 26 at 7:00
  • $\begingroup$ What about the same question asked for the vertices of a regular $3n$-gon ? $\endgroup$
    – Jean Marie
    Apr 26 at 7:33

1 Answer 1


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:

a triangle inscribed in a circle

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:

a triangle in which there's a tie for leftmost

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:

a triangle with the leftmost point colored red and the other two points colored blue

And here's the full coloring:

a circle, one third of which is colored red

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 ^_^

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    $\begingroup$ Or, to put the last paragraph simply, you can color a half-open half-closed arc equal to 1/3 of the circle red and the remaining 2/3 blue. $\endgroup$ Apr 26 at 7:23
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    $\begingroup$ Yes, this answer was super helpful and I've upvoted it! Michal's comment was an excellent addition too. Thank you so much both for the swift answer! I'll wait just a bit before accepting (just in case there are other contributions). $\endgroup$
    – Prism
    Apr 26 at 7:25
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    $\begingroup$ @MichalAdamaszek -- Thanks! I figured there was some easy way to visualize this coloring (I knew you could always select the "leftmost" point in the triangle, once you choose a way to break ties) but I wanted to go to bed, and didn't think hard enough to see how simple it was, haha. In hindsight that's obvious, and I should have probably been able to just see it. If you add it as an answer yourself, I would definitely +1 it (especially if there's a picture). If not, I'll edit my answer sometime tomorrow to include a picture of the "1/3 red" coloring ^_^ $\endgroup$ Apr 26 at 7:30

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