25. For an equilateral triangle $ABC$ with side length 1, each side is divided into $n$ equal segments. Lines parallel to the other two sides are drawn through each division point, dividing the triangle into smaller triangles. The vertices of these smaller triangles are called nodes. A real number is placed at each node. It is known that:
(i) The real numbers placed at points $A, B, C$ are $a, b, c$;
(ii) In each rhombus formed by two adjacent smallest triangles, the sums of the numbers placed at the opposite vertices are equal.
Find: (1) the 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.
(1987 2nd CMO Problem)
25.(1) Consider any trapezoid composed of 3 smallest equilateral triangles in sequence with common sides. As shown in the figure $A_{1} A_{3} A_{4} A_{5}$, let $A_{2}$ be the midpoint of $A_{1} A_{5}$. If the number placed on $A_{i}$ is $a_{i}(i=1,2,3,4,5)$, then $a_{1}+a_{4}=a_{2}+a_{3}, a_{2}+a_{4}=a_{5}+a_{3}$, hence $a_{1}-a_{2}=a_{2}-a_{5}$. The above equation indicates that the 3 numbers placed on the three adjacent nodes $A_{1}, A_{2}, A_{5}$ on the same line form an arithmetic sequence. Since such a trapezoid is arbitrarily chosen, the numbers placed on the nodes on the same line form an arithmetic sequence. Without loss of generality, let $a \leqslant b \leqslant c$, and consider four cases: (i) When $a=b=c$, then the numbers placed on all nodes are equal, so $r=0$; (ii) When $a<b<c$, except for $A$ and $C$, let the number placed on any other node be $x$, then $a<x<c$, so $r=AC=1$; (iii) When $a<b=c$, if $n$ is even, then the midpoint $D$ of $BC$ is a node, and the number placed on $D$ is $\frac{1}{2}(b+c)=c$, and the numbers placed on the other nodes are all greater than $a$ and less than $b$, hence $r=AD=\frac{\sqrt{3}}{2}$; when $n$ is odd, the midpoint $D$ of $BC$ is not a node, and the distance from the node $M$ closest to $D$ to $D$ is $\frac{1}{2n}$, and the number placed on $M$ is $b=c$, hence $r=AM=\sqrt{AD^2+DM^2}=\frac{1}{2n}\sqrt{3n^2+1}$; (iv) When $a=b<c$, by symmetry, the same conclusion as (iii) can be obtained. In summary: When $a=b=c$, $r=0$; when $a<b<c$, $r=1$; when $a<b=c$ or $a=b<c$ and $n$ is even, $r=\frac{\sqrt{3}}{2}$; when $a<b=c$ or $a=b<c$ and $n$ is odd, $r=\frac{1}{2n}\sqrt{3n^2+1}$.
(2) From the proof in (1), it is not difficult to see that if the numbers placed on points $A, B, C$ are $b, c, a$ or $c, a, b$, and the requirement (ii) for the numbers placed on each node is maintained, then the total sum of the numbers placed on all nodes is still $S$. Adding the three numbers placed on the same node in the above three placements results in a new placement method, which still satisfies the requirement (ii), and the numbers placed on $A, B, C$ are all $a+b+c$. From the proof in (1), we know that the numbers placed on each node are all $a+b+c$, hence $3S=(a+b+c) \times (\text{number of nodes})=(a+b+c) \times \frac{1}{2}n(n+1)$, so $S=\frac{1}{6}n(n+1)(a+b+c)$.