2012 AMC 10B Problem 15

Attempt Problem 15 of the 2012 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2012 AMC 10B solutions, or check the answer key.

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15.

In a round-robin tournament with 6 teams, each team plays one game against each other team, and each game results in one team winning and one team losing. At the end of the tournament, the teams are ranked by the number of games won. What is the maximum number of teams that could be tied for the most wins at the end of the tournament?

2 2

3 3

4 4

5 5

6 6

Answer: D
Concepts:graph theoryextremal argument
Difficulty rating: 1540
Solution:

They would have to share (62)=15\binom 62 = 15 wins.

This means we cannot have a 66 way tie as that would be 2.52.5 wins per team.

If we had a 55 way tie, each team could have 33 wins, which is possible if one team loses all of its games, and out of the 55 winning teams, they each split their games.

If we label the teams from 11 to 5,5, and designate team 66 to lose all their games. To get a 55 way tie, we could have each team 1x51\le x \le 5 beat team x+1mod5x+1 \mod 5 and team x+2mod5,x+2 \mod 5, as well as team 6.6.

Thus, the correct answer is D .

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