Given the square $ABCD\ :\ A(0,0),B(a,0), \cdots$ Choose $K(\lambda,0)$ and $L(a,\mu)$. $\triangle KLM$ is equilateral if $\mu=2a-\sqrt{3}(a-\lambda)$ Midpoint of $KL\ :\ x=\frac{\lambda+a}{2}, y=\frac{\mu}{2}$. Eliminating the parameter $\lambda$, we find the locus of the midpoints, part of the line $y=\sqrt{3}x+a(1-\sqrt{3})$.

Let $I$ the midpoint of $KL$ then $MI \perp KL$ thus $MILC$ is cyclic hence $\angle ICB =\angle ICL=\angle IML =30^{\circ }$ therefore the locus is a part of line passing through $C$ with the slope $\frac{\sqrt 3}{3}$ wrt $BC$

an other similar question is to find the locus of their circumcenters

vanstraelen wrote: Given the square $ABCD\ :\ A(0,0),B(a,0), \cdots$ Choose $K(\lambda,0)$ and $L(a,\mu)$. $\triangle KLM$ is equilateral if $\mu=2a-\sqrt{3}(a-\lambda)$ Midpoint of $KL\ :\ x=\frac{\lambda+a}{2}, y=\frac{\mu}{2}$. Eliminating the parameter $\lambda$, we find the locus of the midpoints, part of the line $y=\sqrt{3}x+a(1-\sqrt{3})$. PROF65 wrote: Let $I$ the midpoint of $KL$ then $MI \perp KL$ thus $MILC$ is cyclic hence $\angle ICB =\angle ICL=\angle IML =30^{\circ }$ therefore the locus is a part of line passing through $C$ with the slope $\frac{\sqrt 3}{3}$ wrt $BC$ Both posts are correct, but they are not complete solutions. You still have to find the exact locus of the midpoint. It could be the entire line, or the segment contained in the square, or any other possibility that you haven't discarded yet. A crucial part of the problem (and basically the real problem) is to properly bound the locus.