AIME 1985 · 第 14 题

Solution 1

Let us suppose for convenience that there were $n + 10$ players overall. Among the $n$ players not in the weakest 10 there were $n \choose 2$ games played and thus $n \choose 2$ points earned. By the givens, this means that these $n$ players also earned $n \choose 2$ points against our weakest 10. Now, the 10 weakest players playing amongst themselves played ${10 \choose 2} = 45$ games and so earned 45 points playing each other. Then they also earned 45 points playing against the stronger $n$ players. Since every point earned falls into one of these categories, It follows that the total number of points earned was $2{n \choose 2} + 90 = n^2 - n + 90$. However, there was one point earned per game, and there were a total of ${n + 10 \choose 2} = \frac{(n + 10)(n + 9)}{2}$ games played and thus $\frac{(n + 10)(n + 9)}{2}$ points earned. So we have $n^2 -n + 90 = \frac{(n + 10)(n + 9)}{2}$ so $2n^2 - 2n + 180 = n^2 + 19n + 90$ and $n^2 -21n + 90 = 0$ and $n = 6$ or $n = 15$. Now, note that the top $n$ players got $n(n - 1)$ points in total (by our previous calculation) for an average of $n - 1$, while the bottom 10 got 90 points total, for an average of 9. Thus we must have $n > 10$, so $n = 15$ and the answer is $15 + 10 = \boxed{25}$.

Solution 2

Suppose that there are $n$ players participating in the tournament. We break this up into a group of the weakest ten, and the other $n-10$ people. Note that the $10$ players who played each other generated a total of $\dbinom{10}{2} = 45$ points playing each other. Thus, they earned $45$ playing the $n-10$ other people. Thus, the $n-10$ people earned a total of $10(n-10)-45 = 10n-145$ points playing vs. this group of 10 people, and also earned a total of $10n-145$ playing against themselves. Since each match gives a total of one point, we must have that $\dbinom{n-10}{2}=10n-145$. Expanding and simplifying gives us $n^2-41n+400=0$. Thus, $n=16$ or $n=25$. Note however that if $n=16$, then the strongest $16$ people get a total of $16*10-145=15$ playing against the weakest $10$ who gained $45$ points vs them, which is a contradiction since it must be larger. Thus, $n=\boxed{25}$.

Solution by GameMaster402

Solution 3

Note that the total number of points accumulated must sum to ${p \choose 2} = \frac{p(p-1)}{2}$. Say the number of people is $n$. Consider the number of points gained when the 10 lowest scoring people play each other. The problem tells us that each of these 10 people must earn exactly half of the total number of points they will earn during the whole game. This implies that this group of 10 people must accumulate half their total combined points after they (the 10 people) all play each other, meaning they must earn the other half of their points by playing the $n-10$ stronger players. The problem also tells us that the $n-10$ people who aren't part of the losers group will earn half of their points by playing the $10$ losers. Since the $n-10$ group and $10$ losers will earn half their points by playing each other, the sum of the number of points that they gain playing each other must then be half of the total amount of points earned by everyone in the game. Therefore, $\frac{p(p-1)}{4} = 10(p-10)$. This equation is the same as above, and by the same logic, the answer is $n=\boxed{25}$.

Solution 4 (Expected Value and simple system of equations)

Against any given person player $p_1$ has a $1/3$ chance of obtaining $1$ point, $1/3$ chance of obtaining $0$ points, and $1/3$ chance of obtaining $1/2$ a point. Thus, the expected value of the number of points earned in one game is just $\frac{1}{3} \cdot 1 + \frac{1}{3} \cdot 0 + \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{2}$.

Say there are $n+10$ people. Then, the number of games is determined by $\binom{n+10}{2}$. The expected number of points for player $p_1$ is determined as

$$p_1 = \frac{1}{2} \binom{n+10}{2}$$ Now, let the $n+10$ people be $p_1, p_2, p_3, \dots, p_{n+10}$. The expected number of total points $P$ is determined as the sum of all $p_1, p_2, p_3, \dots, p_{n+10}$, or $\left( \frac{1}{2} \binom{n+10}{2} \right)(n+10)$.

Now we use the other information given. We have that half the total number of points $P$ came from the $10$ people with the least points. WLOG say these people were $p_1, p_2, p_3, \dots, p_{10}$. Thus we have

$$\frac{1}{2} P = 10 \left( \frac{1}{2} \binom{10}{2} \right) + n \left( \frac{1}{2} \binom{n}{2} \right) + 5n$$ If you don't understand this, let's break it down LHS to RHS. On the LHS, it is quite obvious. Half the total points obtain, which is $\frac{1}{2} P$. Now, first term on the RHS. Note what the MAA says in the brackets. The property applies for the 10 lowest. Thus, for each of $p_1, p_2, p_3, \dots, p_{10}$, we must consider their individual chances of winning. Between their respective 10 player bracket, $\binom{10}{2}$ games occur. Then, the expected number of points gained per player is just $\frac{1}{2}$. Combining them together, we obtain the first term. Now, the middle term. We now consider the $n$ people going against each other. This time, we consider each of $p_{11}, p_{12}, p_{13}, \dots, p_{n+10}$, which is $n$ total. Then, there are now $\binom{n}{2}$ games in the bracket, and each player has a $\frac{1}{2}$ chance of winning. Combining all together gives us the middle term. The last term involves each of the $n$ people going against the 10 lowest people. Then, $p_{11}$ goes against each of $p_1, p_2, p_3, \dots, p_{10}$, so does $p_{12}$, $p_{13}$, and all the way until $p_{n+10}$, giving $n$ people against 10 others, or $10n$ games. In each of these games, the player has a $\frac{1}{2}$ expected chance of winning, giving us $5n$. Combining everything together gives us the above equation.

And thus we have the system of equations

$$P = \left( \frac{1}{2} \binom{n+10}{2} \right)(n+10)$$ $$\frac{1}{2} P = 10 \left( \frac{1}{2} \binom{10}{2} \right) + n \left( \frac{1}{2} \binom{n}{2} \right) + 5n$$ Through some simple algebra and combinatorial expansions, we find that the values for $n$ satisfy the equation

$$(n-15)(n-(8+2\sqrt{31}))(n-(8-2\sqrt{31})) = 0$$ The three values of $n$ are

$$n=15, \quad n=8+2\sqrt{31}, \quad n=8-2\sqrt{31}$$ However, we require the number of people, and thus the two latter values are obviously incorrect.

Thus, $n=15$. We had $n+10$ people in the beginning, and thus the number of people at the tournament were $15+10 = \boxed{25}$.

~Pinotation