Shapovalov A.V.

55 boxers participated in a tournament with a "loser leaves" system. The fights proceeded sequentially. It is known that in each match, the number of previous victories of the participants differed by no more than 1. What is the maximum number of fights the tournament winner could have conducted?
We will prove by induction that

a) if the winner has conducted no less than $n$ fights, then the number of participants is no less than $u_{n+2}$;

b) there exists a tournament with $u_{n+2}$ participants, the winner of which has conducted $n$ fights ( $u_{k}-$ Fibonacci numbers).

Base case $\left(n=1, u_{3}=2\right)$ is obvious.

Inductive step. a) Suppose the winner $A$ won the last fight against boxer $B$. The remaining fights effectively split into two tournaments: one won by $A$, and the other by $B$. In the first tournament, the winner $A$ conducted no less than $n-1$ fights, so the number of participants is no less than $u_{n+1}$. In the second tournament, the winner $B$ conducted no less than $n-2$ fights, so the number of participants is no less than $u_{n}$. Therefore, in the original tournament, the number of participants is no less than $u_{n+1} + u_{n} = u_{n+2}$.
b) It is sufficient to combine in the final fight the winner of a tournament with $u_{n+1}$ participants, who won $n-1$ fights, and the winner of a tournament with $u_{n}$ participants, who won $n-2$ fights.

Since $55 = u_{10}$, it follows from this that the answer is.

## Answer

8 fights.