Problem 4 - Algebra for JEE ADVANCED

Question

Ten persons numbered \(1,2, \ldots, 10\) play a chess tournament, each player playing against every other player exactly one game. It is known that no game ends in a draw. If \(w_{1}, w_{2}, \ldots, w_{10}\) are the number of games won by players \(1,2,3, \ldots, 10\), respectively, and \(I_{1}, l_{2}, \ldots, l_{10}\) are the number of games lost by the players \(1,2, \ldots\) 1 \(O\), respectively, then, a. \(\sum w_{l}=\Sigma l_{i}=45\) b. \(w_{i}+1_{i}=9\) c. \(\sum w l_{1}^{2}=81+\Sigma l_{1}^{2}\) d. \(\sum w_{i}^{2}=\sum l_{i}^{2}\)

Step-by-Step Solution

Answer

Correct relationship: \( \sum w_i^2 = \sum l_i^2 \).

Step 1: Determine Total Number of Games

In a tournament where each player plays against every other player exactly one game, we use the formula for combinations to find the total number of games. Here, there are 10 players, so the number of nodes (participants) using the combination formula is \( \binom{10}{2} = \frac{10 \times 9}{2} = 45 \).

Step 2: Verify Total Wins and Losses

Since no game ends in a draw, every game has a winner and a loser. Thus, the sum of all wins \( \sum w_i \) must equal the total number of games, which is 45, and the same goes for the sum of losses: \( \sum l_i = 45 \).

Step 3: Analyze the Relation between Wins and Losses

Each player plays 9 games (against the other 9 players), so the number of wins plus the number of losses for each player is equal to 9: \( w_i + l_i = 9 \) for each player \( i \).

Step 4: Calculate the Sum of Squares Relation

We need to verify the relation \( \sum w_i^2 = \sum l_i^2 \). Note that from Step 3, for each player, \( l_i = 9 - w_i \). So, the sum of the squares can be rewritten: \( \sum (9 - w_i)^2 = \sum l_i^2 \). Upon rearranging and using the expansion \( (9 - w_i)^2 = 81 - 18w_i + w_i^2 \), the equation becomes \( \sum w_i^2 = \sum l_i^2 \) since \( \sum 81 - 18w_i + w_i^2 = 81 \times 10 - 18 \times 45 + \sum w_i^2 = 81 \times 10 - 18 \times 45 + \sum (9 - w_i)^2 \). This equation holds true, hence \( \sum w_i^2 = \sum l_i^2 \).

Key Concepts

Chess Tournament

In a chess tournament where each participant plays against everyone else, it is vital to determine the total number of games. This particular tournament involves 10 players, with each one playing every other player exactly once. To find out how many games will be played in total, we use the concept of combinatorics.
Each match can be seen as a combination of two players, which can be calculated using the formula for combinations: \[ \binom{n}{k} = \frac{n!}{k! (n-k)!} \]where \(n\) is the total number of players, and \(k\) is the number of players per match. Here, \(n = 10\) and \(k = 2\). Applying the formula gives us:

• \(\binom{10}{2} = \frac{10 \times 9}{2} = 45\)

Thus, there will be a total of 45 games played in this tournament. This sets the framework for understanding other tournament dynamics.
The outcome of each game will either result in a win or a loss for the players involved. As no games end in a draw, this simplifies the calculations for total wins and losses.

Game Theory

Game theory helps us analyze strategic interactions, such as a chess tournament. In our given tournament, each player plays 9 games, and every game has a definitive outcome: a win or a loss. This allows us to compute how a player's performance can be evaluated using wins and losses.
For each player, the total number of games played is the sum of their wins \(w_i\) and losses \(l_i\). Hence, \[ w_i + l_i = 9 \] for each player. As we have 10 players, this relationship can also be summarized over the entire tournament as:

• The sum of all wins, \(\sum w_i = 45\), equals the total number of games.
• The same holds true for losses, \(\sum l_i = 45\).

Understanding this simple arithmetic ensures that every player's performance adds up in a zero-sum fashion, meaning the total outcomes across the tournament equal the total number of games.

Sum of Squares

The sum of squares concept is crucial when evaluating elements such as dispersions in tournament results. For the chess tournament, we need to show that the sum of squares of wins is equal to that of losses, expressed mathematically as:\[ \sum w_i^2 = \sum l_i^2 \]Given that each player plays 9 matches, we know from previous discussions that \(l_i = 9 - w_i\).
By substituting and expanding, we find:

• \((9 - w_i)^2 = 81 - 18w_i + w_i^2\)

Thus, the equation becomes:\[ \sum (9 - w_i)^2 = 81 \times 10 - 18 \times 45 + \sum w_i^2 \]Simplifying shows it holds that \(\sum w_i^2 = \sum l_i^2\).
This identity aids in understanding how the wins and losses are spread out among participants. It shows a balance in the tournament structure, where the variance or distribution of outcomes maintains a certain harmony across different metrics of performance.