Problem 1422

1. Understanding the Problem and Initial Setup:
- We are given an equilateral triangle $\triangle ABC$ with side length 1.
- Each side of the triangle is divided into $n$ segments by placing $n-1$ points on each side.
- We draw lines parallel to the sides of the triangle through these points, creating smaller triangles within $\triangle ABC$.
- We need to assign real numbers to the vertices (nodes) of these smaller triangles such that certain conditions are satisfied.

2. Condition Analysis:
- Condition (II) states that for any rhombus formed by two smaller triangles sharing a common side, the sum of the numbers at the vertices on one diagonal equals the sum on the other diagonal.
- This condition can be generalized using vectors and linear combinations.

3. Lemma and Proof:
- Lemma: For any subset of nodes $P_1, P_2, \dots, P_k$ with position vectors $\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_k$ satisfying $\sum_{i=1}^k q_i \mathbf{v}_i = 0$ for some real numbers $q_1, \dots, q_k$ with $\sum_{i=1}^k q_i = 0$, we have $\sum_{i=1}^k q_i f(P_i) = 0$.

- Proof:
- We use induction on $k$.
- For $k = 2$, the condition is trivial.
- Assume it holds for $k$. For $k+1$, consider the points $P_1, P_2, \dots, P_{k+1}$ with $\sum_{i=1}^{k+1} q_i \mathbf{v}_i = 0$ and $\sum_{i=1}^{k+1} q_i = 0$.
- We can eliminate points row by row using the given condition, reducing the problem to a smaller triangle.
- In the reduced case, the condition $x \vec{X} + y \vec{Y} + z \vec{Z} = 0$ with $x + y + z = 0$ implies $x = y = z = 0$ due to linear independence.
- Thus, $x f(X) + y f(Y) + z f(Z) = 0$ is trivial, proving the lemma.

4. Application of the Lemma:
- For any node $R$ on segment $\overline{PQ}$, we have $\vec{R} = \lambda \vec{P} + \mu \vec{Q}$ with $\lambda, \mu > 0$ and $\lambda + \mu = 1$.
- By the lemma, $f(R) = \lambda f(P) + \mu f(Q)$, which lies between $f(P)$ and $f(Q)$.
- Therefore, the numbers assigned to nodes other than $A, B, C$ are strictly between the minimum and maximum of $a, b, c$.

5. Minimum Distance Calculation:
- The minimum distance between the node with the maximal number and the node with the minimal number is the diameter of $\triangle ABC$, which is 1.
- If $a, b, c$ are not distinct, the minimum distance is $\frac{1}{n}$.

6. **Sum of Numbers $S$:**
- $\triangle ABC$ has rotational symmetry around its center $O$.
- For any equilateral $\triangle PQR$ with center $O$, $\vec{P} + \vec{Q} + \vec{R} = \vec{A} + \vec{B} + \vec{C}$.
- By the lemma, $f(P) + f(Q) + f(R) = a + b + c$.
- The total number of nodes $N$ is $\frac{(n+1)(n+2)}{2}$.
- Therefore, $S = N \cdot \frac{a + b + c}{3}$.