Fermat Point?!?

Let $B$ have coordinates $(0, 0)$ , $A$ have coordinates $(0, 1)$ , $C$ have coordinates $(\sqrt{3}, 0)$ , and $M$ have coordinates $(p, q)$ .

Then $S_{\triangle MAB} = \frac{1}{2}p$ , $S_{\triangle MBC} = \frac{\sqrt{3}}{2}q$ , and $S_{\triangle MBC} = \frac{\sqrt{3}}{2} - \frac{1}{2}p - \frac{\sqrt{3}}{2}q$ .

And $\overrightarrow{MA} = (-p, 1 - q)$ , $\overrightarrow{MB} = (-p, -q)$ , and $\overrightarrow{MC} = (\sqrt{3} - p, -q)$ .

So $\overrightarrow{MA} \cdot \overrightarrow{MB} = 2p^2 + 2q^2 - 2q$ , $\overrightarrow{MB} \cdot \overrightarrow{MC} = 2p^2 + 2q^2 - 2\sqrt{3}p$ , and $\overrightarrow{MC} \cdot \overrightarrow{MA} = 2p^2 + 2q^2 - 2\sqrt{3}p - 2q$ .

Substituting into the given equations gives $\frac{\frac{1}{2}p}{2p^2 + 2q^2 - 2q} = \frac{\frac{\sqrt{3}}{2}q}{2p^2 + 2q^2 - 2\sqrt{3}p} = \frac{\frac{\sqrt{3}}{2} - \frac{1}{2}p - \frac{\sqrt{3}}{2}q}{2p^2 + 2q^2 - 2\sqrt{3}p - 2q}$ which solves to $p = \frac{\sqrt{3}}{7}$ and $q = \frac{2}{7}$ .

Then $AM = \frac{2}{\sqrt{7}}$ , $BM = \frac{1}{\sqrt{7}}$ , and $CM = \frac{4}{\sqrt{7}}$ , so that $(AM + BM + CM)^2 = \boxed{7}$ .