## Task 4 - 340834

A chess tournament was held in "rounds". These rounds - unlike the common practice - were set up so that in each round, each participant in the tournament played exactly one game (so an even number of players participated) and that in the entire tournament, each participant was scheduled to play exactly one game against every other participant.

Michael and Robert participated in the tournament for 5 rounds, after which they had to drop out. To not further disrupt the remaining tournament, the games they were supposed to play according to the tournament schedule were simply canceled.

Michael told his friends Herbert and Gerd that, as a result, a total of 38 games were played in the entire tournament (with no other withdrawals). Herbert said, "This number is not possible." Gerd countered, "Yes, it is, and if this number is correct, then Michael's statements uniquely determine whether Michael and Robert played against each other in the tournament."

Investigate whether Herbert's or Gerd's opinion is correct! If Gerd's opinion is correct, did Michael and Robert play against each other in the tournament?
If in addition to Michael and Robert, exactly $2 x$ players participated in the tournament, then each of these players played exactly one match against every other; this amounted to

$$
\frac{1}{2} \cdot 2 x(2 x-1)=x \cdot(2 x-1)
$$

matches.

If Michael and Robert did not play against each other in the tournament, they each played two matches in every one of the first 5 rounds. Together with these 10 matches, a total of

$$
x \cdot(2 x-1)+10=38
$$

matches were played, which is possible; because this equation has (as a check shows) the solution $x=4$. Therefore, Herbert's opinion does not hold.

Furthermore, it holds: If Michael and Robert had played against each other in the tournament, they would have played two matches in each of 4 rounds, and only one match against each other in one round. Then, if the total number of 38 matches is correct, together with these 9 matches, a total of

$$
x \cdot(2 x-1)+9=38
$$

matches would have been played. However, this equation has no natural number $x$ as a solution, as, for example, it follows that for all natural numbers $x \leq 4$, $x(2 x-1)+9 \leq 4 \cdot 7+9=37$ results, and for all $x>5$, $x \cdot(2 x-1)+9 \geq 5 \cdot 9+9=54$.

Thus, the possibility that Michael and Robert played against each other in the tournament is ruled out; i.e., from the given information, it unequivocally follows: They did not play against each other; Gert's opinion is correct.