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I know that the answer is 42/3 = 14 points, or in general for a circle with N points it is N/3, but I don't know why it actually works.

Why is the number of equilateral triangles for a circle with N points simply N/3? What happens when N is not divisible by 3?

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Do the triangles have to have integer coordinates or something like that? (Although that isn't possible ;) – Ragnar Dec 17 '13 at 0:46
The points are on the circumference on the circle. The equilateral triangle is made by choosing 3 points on the circumference. – 1110101001 Dec 17 '13 at 0:50
up vote 2 down vote accepted

Measure the central angles, say, from one of the given points. We have points exactly at angles $360^\circ\cdot \displaystyle\frac kN$ with $k=0,...,N-1$.

If $N$ is not divisible by $3$, then try to prove that the difference of any two angles cannot be $120^\circ$ which would be required for the equilateral triangle.

If $3|N$, then look at the vertex which is on the first third, i.e. with angle in $[0^\circ,\,120^\circ)$. An equilateral triangle must have exactly one vertex on each third of the circle, and one vertex already determines the other two. Now, the vertex on the first third can be any out of those $N/3$.

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So if I understood correctly, each equilateral triangle must have one vertex on each third. Since any point outside of the first third of the circle will give an identical triangle to that formed from a point chosen in the first third, you only need to consider how many possible points are there in the first third, which is what N/3 gives you. Is my understanding right? – 1110101001 Dec 17 '13 at 0:56
Yes, yes..$\,\!$ – Berci Dec 17 '13 at 0:58

Let $x_1,\cdots,x_n$ be the points taken clockwise. If your triangle is $x_i,x_j,x_k$ then the angle at $x_k$ is $\frac{\pi}{n}|i-j|$ (it is easy to see with a picture), so it follows that $n$ is a multiple of $3$. This also shows that your points are evenly spaced (with fixed separation) which implies your claim. This could be useful

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