Fermat's Distance

Rotate triangle $BPC$ through $60^\circ$ about $B$ to form triangle $BQC'$ , so that $BPQ$ and $BCC'$ are equilateral triangles. Then $PQ = PB$ and $QC'=PC$ , so that $PA + PB + PC = AP + PQ + QC' \ge AC'$ , and hence we deduce that $AP + BP + CP \ge AC'$ , and this minimum distance is achieved when $P$ is the first Fermat point of $ABC$ , which is the point where $P$ sutends angle $120^\circ$ at each of the arcs $AB$ , $BC$ and $AC$ . In this case $AC' \; = \; \frac{1}{\sqrt{2}} + \sqrt{2} \times \frac{\sqrt{3}}{2} \; = \; \frac{\sqrt{3}+1}{\sqrt{2}} \; = \; \sqrt{2+\sqrt{3}}$ making the answer $2+3=\boxed{5}$ .