RMO Training: more help from Nordic mathematical contest


32 competitors participate in a tournament. No two of them are equal and in a one against one match the better always wins. (No tie please). Show that the gold, silver and bronze medal can be found in 39 matches.


We begin by determining the gold medallist using classical elimination, where we organize 16 pairs and matches, then 8 matches of the winners, 4 matches of the winners in the second round, then 2-semifinal matches and finally one match making 31 matches altogether.

Now, the second best player must have at some point lost to the best player, and as there were 5 rounds in the elimination, there are 5 candidates for the silver medal. Let C_{i} be the candidate who  lost to the gold medalist in round i. Now, let C_{1} and C_{2} play, the winner play against C_{3}, and so forth. After 4 matches, we know the silver medalist; assume this was C_{k}.

Now, the third best player must have lost against the gold medalist or against C_{k} or both. (If the player had lost to someone else, there would be at least three better players.) Now, C_{k} won k-1 times in the elimination rounds, the 5-k players C_{k+1}\ldots C_{5} and if k is greater than one, one player C_{j} with j<k. So there are either (k-1)+(5-k)=4 or (k-1)+(5-k)+1=5 candidates for the third place. At most 4 matches are again needed to determine the bronze winners.

Cheers to Norway mathematicians!

Nalin Pithwa.

Reference: Nordic mathematical contests, 1987-2009.

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