6. There are 90 players in a league. Each of the 90 players plays exactly one match against each of the other 89 players. Each match ends with a win for one player and a loss for the other player, or with a tie for both players. Each player earns 1 point for a win, 0 points for a loss, and 0.5 points for a tie. After all matches have been played, the points earned by each player are added up. What is the greatest possible number of players whose total score can be at least 54 points?
6. When each of 90 players plays against all of the other 89 players exactly once, the total number of games played is $\frac{90 \times 89}{2}=4005$.
This is because each of the 90 players plays 89 games, giving $90 \times 89$, but each of the games is counted twice in this total, since each game is played by two players, giving $\frac{90 \times 89}{2}$.
Exactly 1 point is awarded in total for each game. This is because the two players either win and lose $(1+0$ points) or both tie ( $0.5+0.5$ points $)$.
Therefore, over 4005 games, a total of 4005 points are awarded to the players in the league. Suppose that, after all games have been played, $n$ players have at least 54 points.
Then $54 n \leq 4005$ since these $n$ players cannot have more points than the total number of points awarded to the entire league.
Since $\frac{4005}{54} \approx 74.17$ and $n$ is an integer, this means that $n \leq 74$. (Note that $74 \times 54=3996$ and $75 \times 54=4050$.)
Suppose that 74 players have at least 54 points.
This accounts for at least $74 \times 54=3996$ of the total of 4005 points.
This leaves at most $4005-3996=9$ points to be distributed among the remaining $90-74=16$ players.
These 16 players play $\frac{16 \times 15}{2}=120$ games amongst themselves (each of 16 players plays the other 15 players) and these games generate 120 points to be distributed among the 16 players. This means that these 16 players have at least 120 points between them, and possibly more if they obtain points when playing the other 54 players.
Therefore, these 16 players have at most 9 points and at least 120 points in total, which cannot happen.
Thus, it cannot be the case that 74 players have at least 54 points.
Suppose that 73 players have at least 54 points.
This accounts for at least $73 \times 54=3942$ of the total of 4005 points.
This leaves at most $4005-3942=63$ points to be distributed among the remaining $90-73=17$ players.
These 17 players play $\frac{17 \times 16}{2}=136$ games amongst themselves.
This means that these 17 players have at least 136 points between them.
Therefore, these 17 players have at most 63 points and at least 136 points in total, which cannot happen.
Thus, it cannot be the case that 73 players have at least 54 points.
Suppose that 72 players have at least 54 points.
This accounts for at least $72 \times 54=3888$ of the total of 4005 points.

This leaves at most $4005-3888=117$ points to be distributed among the remaining $90-72=18$ players.
These 18 players play $\frac{18 \times 17}{2}=153$ games amongst themselves.
This means that these 18 players have at least 153 points between them.
Therefore, these 18 players have at most 117 points and at least 153 points in total, which cannot happen.
Thus, it cannot be the case that 72 players have at least 54 points.
Suppose that 71 players have at least 54 points.
This accounts for at least $71 \times 54=3834$ of the total of 4005 points.
This leaves at most $4005-3834=171$ points to be distributed among the remaining $90-71=19$ players.
These 19 players play $\frac{19 \times 18}{2}=171$ games amongst themselves.
This means that these 19 players have at least 171 points between them.
These point totals match. Therefore, it appears that 71 players could have at least 54 points.
It is in fact possible for this to happen.
Suppose that every game between two of the 71 players ends in a tie, every game between one of the 71 players and one of the 19 players ends in a win for the first player, and every game between two of the 19 players ends in a tie.
In this case, each of the 71 players has 70 ties and 19 wins, for a point total of $(70 \times 0.5)+$ $(19 \times 1)=54$, as required.
ANSWER: 71