A short while ago I came across this quite famous Russian problem which has a very intriguing result, but I myself have made little headway on it thus far:

In a chess tournament there are \( 2n+3 \) competitors. Every competitor plays every other competitor precisely once. No 2 matches can be played simultaneously, and after a competitor plays a match he can not play in any of the next *n* matches. Show that one of the competitors who plays in the opening match will also play in the closing match.

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TopNewestHere's an initial thought. The first 2 players cannot play in the next \(n\) games, which involve \(2n\) more people. Hence, the next game after that must involve the last person, and one of the first 2 players.

Can we continue from here? – Calvin Lin Staff · 2 years, 8 months ago

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