1987 CMO 第 2 题

We are given an equilateral triangle ABC with the length of its side equal to $1$ . There are $n-1$ points on each side of the triangle $ABC$ that equally divide the side into $n$ segments. We draw all possible lines that pass through any two of all those $3(n-1)$ points such that they are parallel to one of three sides of triangle $ABC$ . All such lines divide triangle $ABC$ into some lesser triangles whose vertices are called *nodes*. We assign a real number for each *node* such that the following conditions are satisfied:

(I) real numbers $a,b,c$ are assigned to $A,B,C$ respectively;

(II) for any rhombus that is consisted of two lesser triangles that share a common side, the sum of the numbers of vertices on its one diagonal is equal to that of vertices on the other diagonal.

1) Find the minimum distance between the *node* with the maximal number to the *node* with the minimal number;

2) Denote by $S$ the sum of the numbers of all *nodes*, find $S$ .

我们有一个等边三角形 ABC ，其边长等于 $1$ 。三角形 $ABC$ 的每条边上都有 $n-1$ 个点，将边平均分为 $n$ 段。我们绘制穿过所有 $3(n-1)$ 点中任意两个点的所有可能的直线，使它们平行于三角形 $ABC$ 的三边之一。所有这些线将三角形 $ABC$ 分成一些较小的三角形，其顶点称为*节点*。我们为每个*节点*分配一个实数，以满足以下条件：

(i) 将实数$a,b,c$分别赋值给$A,B,C$；

(II) 对于由两个同一条小三角形组成的菱形，其一条对角线上的顶点数之和等于另一条对角线上的顶点数之和。

1) 求编号最大的*节点*到编号最小的*节点*之间的最小距离；

2) 用$S$表示所有*节点*的数量之和，求$S$。