# Round-robin tournament - sum of squares of wins and losses

I was given this problem and I think I solved it but I wanted to double-check my solution using the collective wisdom here. Here is the problem:

Twelve teams play in a round-robin tournament. Each match results in a win or loss (no ties). If the team $n$ wins $a_n$ matches and loses $b_n$ matches, prove that $$\sum_{n=1}^{12} a_n^2 = \sum_{n=1}^{12}{b_n^2}$$

My work: the number of all matches is 66. So the total number of wins and losses will be 66. $$\sum_{n=1}^{12} a_n = \sum_{n=1}^{12}{b_n}=66$$ It's also easy to see that $b_n=11-a_n$, thus $$\sum_{n=1}^{12}{b_n^2}=\sum_{n=1}^{12}{(121 -22a_n +a_n^2)}=1452 - 22\sum_{n=1}^{12}{a_n} + \sum_{n=1}^{12}{a_n^2}= \sum_{n=1}^{12}{a_n^2}$$

Is there an easier solution? Thanks in advance!

• Looks right to me. Aug 25 '17 at 20:50