In the final phase of a television game show, the $N$ finalists competed exactly once with each other, receiving $10$ points for each win, $5$ points for each draw, and no points for losses.
At the final ranking, the total number of points won by each of the players in positions $1$ to $N-10,$ were received as follows: $50\%$ from games played against the players in positions $N-9$ to $N$ and the rest of them in games with all others.
Similarly, $50\%$ of points won by each of the players in positions $N-9$ to $N,$ were received from games against the players that were ranked in positions $1$ to $N-10.$
Find the number of players.
My humble attempt:
All games played were $N(N-1)/2.$
Assuming there were $W$ wins, there were equally $L$ losses and $W=L,$ so there were $N(N-1)/2-2W$ draws.
The total points won by all players were $10W+5(N(N-1)/2-2W) = 5(N(N-1)/2).$
Total points won by each of the players in positions $1$ to $N-10:$
Each player played $N-1$ games, of which $10$ were with the players in the last $10$ positions (clearly different players from them). So he gained $50\%$ of his points from $N-1-10$ games and $50\%$ of his points from $10$ games... and this is where I am stuck!
Any help is highly appreciated!
Call a player "bad" if they are in the bottom $10$, and "good" otherwise. Let $k=N-10$ be the number of good players. Divide the games up into three categories:
As each game played generates a total of $10$ points, we know that the type 1 games generate a total of $450$ points. So the bad players must also earn a total of $450$ points in games of type 2. Similarly, the good players must earn a total of $10\binom{k}{2}=5k(k-1)$ points in games of type 2. But the total number of points earned in type-2 games is $100k$, so we must have $$ 100k=5k(k-1)+450 $$ which reduces to $$ k^2-21k+90=0 $$ This quadratic has solutions $k=15$ and $k=6$. So these are the only two values for $k$ we need to consider.
Now, I claim that we must have $k>10$. To see this, note that the total number of points earned by good players is $ 10k(k-1) $, and so the average score of the good players is $10(k-1)$. Similarly, the total number of points earned by the bad players is $900$, which means their average score is $90$. But each good player individually outscores each bad player. So in particular the average score of the good players must be greater than the average score of the bad players, which implies that $10(k-1)>90$ or $k>10$. So the only possible solution is $k=15$ (and $N=25$).
In order to complete the solution, we must find an example tournament with $k=15$ which satisfies the given property (otherwise, there might be no possible value of $k$ at all!)
We will generate such an example as follows. Name the good players $G_1$ through $G_{15}$ and the bad players $B_1$ through $B_{10}$. Let every game between two good players, or between two bad players, be a draw. Among games between a good player and a bad player, we will say that player $G_m$ beats player $B_n$ if $m-n \equiv \pm 2 \pmod{5}$, and the game is a draw otherwise. (So bad players never beat good players.) For example, player $G_1$ beats players $$ B_3, B_4, B_8, B_9 $$ and draws with everyone else, and player $B_1$ loses to players $$ G_3, G_4, G_8, G_9, G_{13},G_{14} $$ and draws with everyone else.
Now:
As you can see, each player wins exactly half of their points against players of each category. Also, the good players all really do score more points than the bad players. So the conditions of the problem are satisfied for this tournament, meaning that we can have $N=25$ (and cannot have any other value of $N$).