$n$ teams participated in a basketball tournament. Each team has played with each team exactly one game. There was no tie. If in the end of the tournament the $i$-th team has $x_{i}$ wins and $y_{i}$ loses $(1\leq i \leq n)$ prove that: $\sum_{i=1}^{n} {x_{i}}^2=\sum_{i=1}^{n} {y_{i}}^2$
Problem
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Tags: combinatorics
16.07.2017 14:53
$x_i + y_i$ gives the total number of games played by the $i$-th team. Hence it is a constant $= n - 1$ We have, $\sum_{i = 1}^{n}(x_i^2 - y_i^2) = 0$ $\sum_{i = 1}^{n}(x_i + y_i)(x_i - y_i) = 0$ Since $x_i + y_i$ is constant, we must have $\sum_{i = 1}^{n}x_i - \sum_{i = 1}^{n}y_ i = 0$ which is true because total number of games lost must be equal to total number of games won.
14.08.2019 21:16
This is also in Putnam & Beyond - task 888. Let me remind you this book was published in 2007.
15.08.2019 13:05
Duarti wrote: $n$ teams participated in a basketball tournament. Each team has played with each team exactly one game. There was no tie. If in the end of the tournament the $i$-th team has $x_{i}$ wins and $y_{i}$ loses $(1\leq i \leq n)$ prove that: $\sum_{i=1}^{n} {x_{i}}^2=\sum_{i=1}^{n} {y_{i}}^2$ Cauchy Schwarz