1. In an international chess tournament, a win earns 1 point, a draw earns 0.5 points, and a loss earns 0 points. Now, there are 8 players in a round-robin tournament (each pair of players plays one game). After the tournament, it is found that all players have different scores. When the players are ranked by their scores from highest to lowest, the fourth-place player scores 4.5 points, and the second-place player's score equals the total score of the last four players. How many points did the top three players score? Explain your reasoning.
(1998, Beijing Middle School Mathematics Competition Junior Division)
Solution: Let the scores of the top three players be $a_{1}, a_{2}, a_{3}$ $\left(a_{1}>a_{2}>a_{3}>4.5\right)$.

Since 8 players only participate in 7 matches, the maximum score is 7, i.e., $a_{1} \leqslant 7$. The total number of matches played is $\frac{7 \times 8}{2}=$ 28 matches, with a total score of 28 points. Also, the score of the second player is the sum of the scores of the last four players, so,
$$
a_{1}+2 a_{2}+a_{3}=28-4.5=23.5 \text {. }
$$

Since, in each match, the scores can only be $0, 0.5, 1$, and $2 a_{2}$ is an integer, $a_{1}$ and $a_{3}$ must be one integer and one decimal.
Since, $7 \geqslant a_{1}>a_{2}=a_{3}>4.5$, so,
$4 a_{1}>23.5,4 a_{3} \geqslant 5$.
Therefore, $a_{3}$ can only be $5.5$ or $5$.
When $a_{3}=5.5$, because $a_{1}>a_{2}>a_{3}$, $a_{1}$ can only be 7, at this time $2 a_{2}=11, a_{2}=5.5=a_{3}$, which is incorrect.
When $a_{3}=5$, $a_{1}$ must be 6.5, at this time $2 a_{2}=$ $12, a_{2}=6$, which meets the requirements.
From the above, the scores of the top three are $6.5,6,5$.