 skylsn

2022-06-25

In a chess tournament, each pair of players plays exactly one game. No game is drawn. Suppose the ${i}^{th}$ player wins ${a}_{i}$ games and loses ${b}_{i}$ games. Show that
$\sum {a}_{i}^{2}=\sum {b}_{i}^{2}$
$\sum {a}_{i}^{2}=\sum {b}_{i}^{2}$ lodosr

Expert

$\sum \left({a}_{i}^{2}-{b}_{i}^{2}\right)=\sum \left({a}_{i}+{b}_{i}\right)\left({a}_{i}-{b}_{i}\right)=\left(n-1\right)\sum \left({a}_{i}-{b}_{i}\right)$
$=\left(n-1\right)\left(\sum {a}_{i}-\sum {b}_{i}\right)=\left(n-1\right)\left(\left(\genfrac{}{}{0}{}{n}{2}\right)-\left(\genfrac{}{}{0}{}{n}{2}\right)\right)=0,$
where $n$ is the total number of players. (Each player plays $n-1$ games, so ${a}_{i}+{b}_{i}=n-1$, and there are a total of $\left(\genfrac{}{}{0}{}{n}{2}\right)$ games played, so $\sum {a}_{i}=\sum {b}_{i}=\left(\genfrac{}{}{0}{}{n}{2}\right)$ Gybrisysmemiau7

Expert

Consider flipping the result of one game, say between player 1 and 2.
Then $\sum \left({a}_{i}{\right)}^{2}-\sum \left({b}_{i}{\right)}^{2}$ changes by $2\left({a}_{1}+{b}_{1}\right)-2\left({a}_{2}+{b}_{2}\right)=0$ as ${a}_{i}+{b}_{i}=n-1$, where $n$ is the number of players.
Thus $\sum \left({a}_{i}{\right)}^{2}-\sum \left({b}_{i}{\right)}^{2}$ is the same no matter what the results and thus is same as the extreme case when the winner wins all matches, the runner up all but one etc, for which the sum is $0$.

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