, 2005

Solution:

Part a): For reasons of symmetry, it suffices first to find the most favorable network of segments for the points $O(0,0)$, $A(1,0)$ and $C(0,1)$. By reflection through $O$ one then obtains the desired network for the points of $M$. By a rotation of the plane $xOy$ about $O$ through an angle of $60^{\circ}$ (Fig. 1), the triangle $\triangle OAC$ is carried into the triangle $\triangle OA' C'$, where the segment $P'C'$ is the image of $PC$. The shortest system of segments connecting the points $O$, $A$ and $C$ as required therefore has a total length equal to that of the segment $AC'$.

Fig. 1

In the triangle $\triangle OAC'$, however, $AC' = \sqrt{2-2 \cos 150^{\circ}} = \sqrt{2+\sqrt{3}}$ (law of cosines). By reflection through the origin $O$ one obtains, for part a), a shortest system of segments with total length $2 \sqrt{2+\sqrt{3}} = \sqrt{2} + \sqrt{6}$.

Part b): Likewise, for reasons of symmetry, it suffices first to find the shortest system of segments for the points $D(-1, 1)$, $O(0, 0)$, $A(1, 0)$, $B(1, 1)$ and $C(0, 1)$. Since the distance from $D$ to the remaining four points is at least $1$, such a system has a total length of $1 + \frac{\sqrt{2} + \sqrt{6}}{\sqrt{2}} = 1 + 1 + \sqrt{3} = 2 + \sqrt{3}$ (see Fig. 2).

Fig. 2

Thus (after reflection through $O$) one obtains for b) a total length of $2(2+\sqrt{3}) = 4 + 2\sqrt{3}$.