|June 15th, 2011, 11:19 AM||#1|
Joined: Jun 2010
10 CHESS-PLAYERS are in a championships
10 CHESS-PLAYERS are in a championship, Each chessplayer played (are 45 games in totals) with others only one times,each winner of the game win 1 point, and each losser win -1 point, In the end of the championship are at least 32 games from 45 that has no winner and no looser,
Prove that there exist at least 2 player with equal points. Thank you
|June 15th, 2011, 11:44 AM||#2|
Joined: Nov 2006
From: UTC -5
Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms
Re: 10 CHESS-PLAYERS are in a championships
So there were only 13 games that resulted in points?
If no one ended up at zero, then that would require at least 1+2+3+4+5 + 1+2+3+4+5 = 30 positive or negative points, and we only have 26 = 13*2. So someone had to have zero; ignore that person and focus on the remaining nine.
Of the nine, either five or more had positive scores or five or more had negative scores. Suppose WLOG that five or more had positive scores. Each person scored 1 or more; at least four people scored 2 or more (else they would have the same score as the lowest-scoring positive person); at least three scored 3 or more; etc. for a total of at least 1+2+3+4+5 = 15 points. But there were only 15 games, QED.
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