Problem 1462

1. Graph Interpretation: Consider a graph $G$ where each vertex represents a contestant and each edge represents a match between two contestants. The edge is colored based on the score of the loser in the match. There are 11 possible colors (0 through 10).

2. Existence of a Vertex with Multiple Colors: Choose a vertex $v$ in $G$ such that $v$ has neighbors of at least two different colors. Such a vertex must exist because if all edges were of the same color, it would contradict the problem statement that there are matches with scores 11-0 and 11-10.

3. Application of the Pigeonhole Principle: By the Pigeonhole Principle, there is a color $c$ such that $v$ has at least 10 neighbors connected by edges of color $c$. Let $X$ be the set of these neighbors.

4. Finding a Suitable Trio:
- If there are two vertices in $X$ connected by an edge of a color different from $c$, then we can choose these two vertices and $v$ to form the required trio.
- Assume all edges between vertices in $X$ are of color $c$.

5. Considering Another Color: Consider another color $c'$ such that there is at least one edge from $v$ of color $c'$. Let $v'$ be a vertex such that the edge between $v$ and $v'$ is of color $c'$.

6. **Checking Edges from $v'$ to $X$**:
- For each vertex $v_0 \in X$, if the edge between $v_0$ and $v'$ is of color $c$ or $c'$, then we can choose $v, v', v_0$ to form the required trio.
- Otherwise, all edges between vertices in $X$ and $v'$ must be of a color different from $c$ and $c'$.

7. Application of the Pigeonhole Principle Again: Since there are 9 remaining colors and $|X| \geq 10$, by the Pigeonhole Principle, there must be at least two vertices $v_1, v_2 \in X$ such that the edges between $v_1$ and $v'$, and $v_2$ and $v'$ are of the same color, distinct from $c$ and $c'$.

8. Conclusion: We can choose $v_1, v_2$, and $v'$ to form the required trio. The edges between $v_1$ and $v_2$ are of color $c$, and the edges between $v_1$ and $v'$, and $v_2$ and $v'$ are of the same color, distinct from $c$.

$\blacksquare$