Saudi Arabian Mathematical Competitions

We will prove the statement by induction on the maximum number of members in clubs $A, B, C$.

For $n=1$, each club has exactly one member and the statement is obviously true.

Suppose that when the maximum numbers in three clubs is $n \geq 1$, then one of three above conditions holds. Assume that $a \in A$ is friend of all members in $B$. Consider a new member $x$ to make the maximum number of members increased by 1. It is easy to see that if $x \in A$ or $x \in C$, then the conditions are still true (since the members in $B$ remains unchanged).

If $x \in B$, denote $B=\{b_{1}, b_{2}, \ldots, b_{n}, b_{n+1}\}$ with $b_{n+1} \equiv x$. In case $(a, x)$ are friend, the condition 1) is true and we are done. So we may assume that $(a, x)$ are not friend. Partition $C$ into two subsets, $C_{1}$ has members that are friend of $a$, and $C_{2}$ has members that are not friend of $a$. Take $c_{1} \in C_{1}$ and $c_{2} \in C_{2}$ (if any).

For $1 \leq k \leq n$, consider $(a, b_{k}, c_{1})$, we have $(a, b_{k})$ are friend (since before adding $x$, member $a$ is friend of all members of $B$) and $(a, c_{1})$ are friend so $(b_{k}, c_{1})$ are not friend (by the given condition).

Consider $(a, x, c_{2})$, we have $(x, c_{2})$ are friend (since $(a, x), (a, c_{2})$ are not friend).

- If $x$ is friend of all members in $C_{1}$ then $x$ is friend of all members in $C = C_{1} \cup C_{2}$ and condition 2) is true.
- Otherwise, $\exists c_{1} \in C$ such that $c_{1}$ is not friend of $x$, also is not friend of all members in $B$. If $\exists a' \in A$ such that $(a', c_{1})$ are not friend, then consider $(a', b_{k}, c_{1})$ with $1 \leq k \leq n+1$ then $a'$ is friend of all $b_{k}$, condition 1) is true. Otherwise, $c_{1}$ are friend of all members in $A$ and condition 3) is true.

Hence, in all cases, the condition is always true.