FINAL ROUND

If $A$ lies under the line $\ell$, then the required locus is the line $BC$, where $B\in \ell$, $C\in m$, $m\parallel \ell$, $m\nparallel A$, and the triangle $ABC$ is an equilateral triangle.

Without loss of generality suppose that $A$ lies under the line $\ell$. Show that the required locus is the line $BC$, where $B\in \ell$, $C\in m$, $m\parallel \ell$, $m\nparallel A$, and the triangle $ABC$ is an equilateral triangle.

First show that all points $N$ lie on $BC$. If $M$ coincides with $B$, then it is evident that $N$ coincides with $C$. Let $M$ be an arbitrary point of $\ell$ different from $B$. In the right half-plane with respect to the line $AM$ we construct the ray with the origin at $A$ such that the angle between this ray and the line $AM$ is equal to $60^{\circ }$. Let $K$ be the point of intersection of this ray and the line $BC$ (see the Fig.). Let $D$ be the point on the line $\ell$ (at the left of $B$) such that the triangle $ABD$ is equilateral. Note that $$\begin{array}{rl}\angle MAD& =\angle MAK-\angle DAK=60^{\circ }-\angle DAK=\\ & =\angle DAB-\angle DAK=\angle KAB.\end{array}(1)$$

Moreover, $$\begin{array}{rl}\angle MDA& =180^{\circ }-\angle BDA=180^{\circ }-60^{\circ }=120^{\circ }=\\ & =180^{\circ }-\angle ABC=\angle KBA.\end{array}(2)$$ By construction, $AD=AB$, so from (1) and (2) it follows that the triangles $MAD$ and $KAB$ are equal. Therefore, $AM=AK$, and since, by construction, $\angle MAK=60^{\circ }$, we see that the triangle $MAK$ is equilateral, i.e., $K$ coincides with $N$, which yields $N\in BC$.

It remains to show that for any point $N$, $N\in BC$ there exists a point $M\in l$ such that the triangle $NAM$ is an equilateral triangle.

Indeed, let $N\in BC$. In the left half-plane with respect to the line $AN$ we construct the ray with the origin at $A$ such that the angle between this ray and the line $AN$ is equal to $60^{\circ }$. Let $M$ be the point of intersection of this ray and the line $l$ (see the Fig.). Let $D$ be the point on the line $l$ (at the left of $B$) such that the triangle $ABD$ is equilateral. Note that $$\begin{array}{rl}\angle MAD& =\angle MAN-\angle DAN=60^{\circ }-\angle DAN=\\ & =\angle DAB-\angle DAN=\angle NAB\end{array}(3)$$

Moreover, $$\begin{array}{rl}\angle MDA& =180^{\circ }-\angle BDA=180^{\circ }-60^{\circ }=120^{\circ }=\\ & =180^{\circ }-\angle ABC=\angle NBA.\end{array}(4)$$ By construction, $AD=AB$, so from (3) and (4) it follows that the triangles $MAD$ and $NAB$ are equal. Therefore, $AM=AN$, and since, by construction, $\angle MAN=60^{\circ }$, we see that the triangle $MAN$ is equilateral, as required.