## Task 4 - 060914

In a chess tournament with 8 participants, each player played exactly one game against every other player. At the end of the tournament, all participants had different point totals. The player in second place won exactly as many points as the last four players combined. For a win, a player received 1 point, for a draw, $\frac{1}{2}$ point, and for a loss, no points.

How did the game between the players who finished in 4th and 6th place end?
Each participant played exactly 7 games and could achieve a maximum of 7 points (if they won all games).

The four chess players who took the last four places had to play exactly 6 games among themselves. The 6 points to be distributed were thus divided among them.

Since the player who took second place, according to the problem, won exactly as many points as the last four together, he must have achieved at least 6 points. However, he could not have achieved more than 6 points; for if he defeated the first place player in addition to the others, he would have been first, and if he drew against this player (while winning against all other players), the first and second place players would have the same number of points, contrary to the conditions.

Thus, the last four players must have achieved exactly 6 points together, meaning they lost all their games against the first four players. Consequently, the fourth player also defeated the sixth player.

Moreover, all conditions of the problem are consistent with each other, as shown by the following scheme:

|  | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| 1 |  | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
| 2 | 0 |  | 1 | 1 | 1 | 1 | 1 | 1 |
| 3 | 0 | 0 |  | 1 | 1 | 1 | 1 | 1 |
| 4 | 0 | 0 | 0 |  | 1 | 1 | 1 | 1 |
| 5 | 0 | 0 | 0 | 0 |  | 1 | 1 | 1 |
| 6 | 0 | 0 | 0 | 0 | 0 |  | 1 | 1 |
| 7 | 0 | 0 | 0 | 0 | 0 | 0 |  | 1 |
| 8 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |  |

### 6.8.2 II. Round 1966, Class 9