ii. For an equilateral triangle $ABC$ with side length 1, each side is divided into $n$ equal segments, and lines parallel to the other two sides are drawn through each division point, dividing the triangle into smaller triangles. The vertices of these small triangles are called nodes. A real number is placed at each node. It is known that:
(i) The numbers placed at points $A, B, C$ are $a, b, c$ respectively;
(ii) In each rhombus formed by two adjacent smallest triangles sharing a common side, the sums of the numbers placed at the two pairs of opposite vertices are equal.
Find:
(1) The shortest distance $r$ between the node with the maximum number and the node with the minimum number;
(2) The sum $S$ of all numbers placed at the nodes.
(1) Consider a trapezoid composed of three small triangles with common sides, as shown in the trapezoid $A_{1} A_{2} A_{4} A_{5}$ in Figure 2 (Figure 2 is the case when $n=5$). Here, $A_{3}$ is the midpoint of $A_{1} A_{5}$. Suppose $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ are placed on $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}$, respectively. By the problem's conditions, $a_{1} + a_{4} = a_{2} + a_{3}$; $a_{3} + a_{4} = a_{2} + a_{5}$. Subtracting these two equations yields $a_{1} - a_{3} = a_{3} - a_{5}$. This indicates that the numbers placed on the three adjacent nodes $A_{1}, A_{3}, A_{5}$ on the same line form an arithmetic sequence. Since such a trapezoid can be chosen arbitrarily, this proves that the numbers placed on nodes on the same line form an arithmetic sequence. Therefore, if $a, b, c$ are all distinct, then $r=1$; if two of $a, b, c$ are the same, when $n$ is even, $r=\frac{\sqrt{3}}{2}$, and when $n$ is odd, $r=\frac{1}{2 n} \sqrt{3 n^{2}+1}$; when $a=b=c$, $r=0$.
(ii) $S=\frac{1}{6} n(n+1)(a+b+c)$. A simple method to find this is to use symmetry. In fact, it is not difficult to see that if $b, c, a$ or $c, a, b$ are placed on $A, B, C$ respectively, and property (ii) is still retained, then the total sum of the numbers placed on all nodes remains $S$. Now, if we add the numbers placed on the same node in the three placements, we get a new placement method. In this new placement, the numbers placed on $A, B, D$ are all $(a+b+c)$, and this placement clearly still satisfies property (ii). From the proof of (i), we know that the number placed on each node is $(a+b+c)$, so $3 S = (a+b+c) \times$ number of nodes $= \frac{1}{2} n(n+1) \cdot (a+b+c)$, thus $S = \frac{1}{6} n(n+1)(a+b+c)$.