Define $\epsilon = 10^{-2017}$. Let us take the vertices of the equilateral triangle to be $(0, 0), (2a, 2b), (a - \sqrt{3}b, \sqrt{3}a + b)$. Then it suffices to have $$a^2 + b^2 = \frac{l^2}{4} := x$$and $||a||, ||b||, ||\sqrt{3}a||, ||\sqrt{3}b|| < \frac{\epsilon}{4}$. Let $S$ denote the set of non-negative integers $m$ with $||\sqrt{3}m|| < \frac{\epsilon}{12}$. We claim that for sufficiently large $x$, there exist $m, n \in S$ and $\alpha \in [0, \frac{\epsilon}{12}]$ such that $(m+\alpha)^2 + n^2 = x$. Once this is proved, we can set $a = m + \alpha$ and $b = n$ to obtain the desired equilateral triangle. To prove the preceding claim, we need a lemma: Lemma. There exists a positive integer $L$, such that among every $L$ consecutive non-negative integers, there is at least one that belongs to $S$. This is a corollary of Dirichlet's approximation theorem, so we omit the proof. Fix any sufficiently large $x$. Take $m$ to be the largest integer in $S$ not exceeding $\sqrt{x}$, then by Lemma $m \geq \sqrt{x} - L$ and so $x - m^2 \leq 2L \sqrt{x}$. Next take $n$ to be the largest integer in $S$ not exceeding $x - m^2$, then we can similarly deduce $$0 \leq x - m^2 - n^2 \leq 2L \sqrt{2L\sqrt{x}} < 4L^2 x^{\frac{1}{4}}.$$If $\alpha$ is such that $x = (m+\alpha)^2 + n^2$, then the leftmost inequality above gives $\alpha \geq 0$. Moreover, from $2m \alpha < 2m \alpha + \alpha^2 = x - m^2 - n^2 < 4L^2 x^{\frac{1}{4}}$ and $m > \sqrt{x} - L$, we can deduce $\alpha \leq \frac{\epsilon}{12}$. This completes the proof.

Well, in our winter camp I heard that this is possible with $ n- $ polygons too. I managed to solve the problem in the contest but too much little time was left to elaborate it all

toto1234567890 wrote: Well, in our winter camp I heard that this is possible with $ n- $ polygons too The same argument works, does it not ? We just need to use Kronecker's simultaneous approximation theorem instead of Dirichlet's theorem.

Yeah, that's right. I think the educators thought the students knew only Dirichlet's