This problem is from the Test of Mathematics, TOMATO Subjective Problem no. 172 based on the Round Robin tournament.
Problem : Suppose there are [latex] {k}[/latex] teams playing a round robin tournament; that is, each team plays against all the other teams and no game ends in a draw.Suppose the [latex] {i^{th}}[/latex] team loses [latex] {l_{i}}[/latex] games and wins [latex] {w_{i}}[/latex] games. Show that
[latex] {{\displaystyle}{\sum_{i=1}^{k}{l_i^{2}}}}[/latex] = [latex] {{\displaystyle}{\sum_{i=1}^{k}{w_i^{2}}}}[/latex]
Solution : Each team plays exactly one match against each other team.
Consider the expression [latex] \displaystyle{\sum_{i=1}^{k} l_i^{2} - {w_i^2} = \sum_{i=1}^{k}(l_i + w_i)(l_i - w_i) } [/latex]
Since each team plays exactly k-1 matches and no match ends in a draw, hence number of wins plus numbers of loses of a particular team is k-1 (that is the number of matches it has played). In other words [latex] l_i + w_i = k-1 [/latex] for all i (from 1 to k).
Hence
[latex] \displaystyle{\sum_{i=1}^{k} l_i^{2} - {w_i^2} } [/latex]
[latex] \displaystyle{= \sum_{i=1}^{k}(l_i + w_i)(l_i - w_i) } [/latex]
[latex] \displaystyle{= \sum_{i=1}^{k}(k-1)(l_i - w_i) } [/latex]
[latex] \displaystyle{= (k-1)\left( \sum_{i=1}^{k} l_i - \sum_{i=1}^{k} w_i\right) } [/latex]
But [latex] \displaystyle{ \sum_{i=1}^{k} l_i = \sum_{i=1}^{k} w_i } [/latex] (as total number of loses = total number of matches = total number of wins; as each match results in a win or lose of some one)
Hence [latex] \displaystyle{= (k-1)\left( \sum_{i=1}^{k} l_i - \sum_{i=1}^{k} w_i\right) = (k-1) \times 0 = 0 } [/latex]
Therefore [latex] \displaystyle{\sum_{i=1}^{k} l_i^{2} - {w_i^2} = 0 } [/latex] implying [latex] {{\displaystyle}{\sum_{i=1}^{k}{l_i^{2}}}}[/latex] = [latex] {{\displaystyle}{\sum_{i=1}^{k}{w_i^{2}}}}[/latex]
Proved.
