RMO Training: more help from Nordic mathematical contest

Problem:

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.

Solution:

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.

Amazon India link:

https://www.amazon.in/Nordic-Mathematical-Contest-1987-2009-Todev/dp/1450519830/ref=sr_1_1?s=books&ie=UTF8&qid=1518386661&sr=1-1&keywords=Nordic+mathematical+contest