## Task 3 - 201223

In a football tournament, $n$ teams participated. Each team played against every other team exactly once.

The winning team received 2 points, the losing team received no points, and in the event of a draw, both teams received one point each.

After the tournament, the teams were ranked 1, 2, .., n in the final table according to their total points in descending order. (In the event of a tie, additional criteria were used to distinguish between teams.)

Determine the largest possible number that can occur as the point difference between two directly adjacent teams in the final table in all (under these rules) possible tournaments.
For every possible tournament and for any two teams occupying adjacent positions $k$ and $k+1 (1 \leq k \leq n-1)$ in the final standings, the following holds:

The teams occupying positions $1, \ldots, k$ have played $\frac{k(k-1)}{2}$ games among themselves and have thus earned a total of $2 \cdot \frac{k(k-1)}{2} = k(k-1)$ points. Additionally, these teams have played a total of $k \cdot (n-k)$ games against the teams occupying positions $k+1, k+2, \ldots, n$ and have earned at most $2 k \cdot (n-k)$ points in these games.

Thus, the teams occupying positions $1$ to $k$ have earned no more than

$$
k(k-1) + 2 k(n-k) = k(2n - k - 1)
$$

points in total.

From this, it follows that the team in position $k$ has not earned more than $2n - k - 1$ points; for if this were the case, then each of the teams in positions $1, \ldots, k$ would also have earned more than $2n - k - 1$ points, resulting in a total of more than $k(2n - k - 1)$ points for these teams.

The $n-k$ teams in positions $k+1, \ldots, n$ have played a total of $\frac{(n-k)(n-k-1)}{2}$ games among themselves and have thus earned a total of $(n-k)(n-k-1)$ points.

Therefore, the team in position $k+1$ must have earned at least $n - k - 1$ points; for if it were less, then the same would be true for each of the teams in positions $k+1, \ldots, n$, and the total points from these games would be less than $(n-k)(n-k-1)$.

Consequently, in any possible tournament, the point difference between any two teams occupying adjacent positions $k$ and $k+1$ in the final standings cannot exceed

$$
(2n - k - 1) - (n - k - 1) = n
$$

If an example of a tournament is provided where two adjacent teams have a point difference of $n$, then $n$ is proven to be the desired number.

Such an example is obtained if, in a tournament, the team in position 1 wins against all other teams, thus earning $2n - 2$ points, and the other teams play each other to a draw, so that each of these teams earns $n - 2$ points for these $n - 2$ games and thus a total of $n - 2$ points in the final standings.

The point difference between the teams in positions 1 and 2 is then $(2n - 2) - (n - 2) = n$ points.

Adapted from $[5]$