Given $\triangle ABC\ :\ A(0,0),B(b,\sqrt{3}b),C(c,0)$; choose $M(p,q)$. Points $B'(p,0),C'(\frac{p+\sqrt{3}q}{4},\frac{\sqrt{3}(p+\sqrt{3}q)}{4}\ )$ and $A'(\frac{b^{2}(3c+p+\sqrt{3}q)-bc(2p+\sqrt{3}q)+c^{2}p}{4b^{2}-2bc+c^{2}},\frac{\sqrt{3}b[c(c-p)-b(c-p-\sqrt{3}q)]}{4b^{2}-2bc+c^{2}}\ )$. The sum $A'B+B'C+C'A$ is constant: $\frac{4b^{2}-b(c+p+\sqrt{3}q)+cp}{\sqrt{4b^{2}-2bc+c^{2}}}+c-p+\frac{p+\sqrt{3}q}{2}=t$. Let $\sqrt{4b^{2}-2bc+c^{2}}=w$. The locus of the points $M$ is a line with equation: $q=\frac{\sqrt{3}p[w+2(b-c)]-2\sqrt{3}[(c-t)w+b(4b-c)]}{3(w-2b)}$.