Maths Olympiad Prep

Track / Stage 3 / 69 of 260 #69 of 1964

Problem 69

AMC 10/12, early questions
Combinatorics Difficulty 3.1 Multiple choice

In a tournament there are six teams that play each other twice. A team earns 33 points for a win, 11 point for a draw, and 00 points for a loss. After all the games have been played it turns out that the top three teams earned the same number of total points. What is the greatest possible number of total points for each of the top three teams?

Pick one

Official solution

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to another. This gives equality, as each team wins once and loses once as well. For a win, we have 33 points, so a team gets 3×2=63\times2=6 points if they each win a game and lose a game. This case brings a total of 18+6=2418+6=24 points.
Therefore, we use Case 2 since it brings the greater amount of points, or (C) 24\boxed {\textbf {(C) }24}.

Note that case 2 can be easily seen to be better as follows. Let xAx_A be the number of points AA gets, xBx_B be the number of points BB gets, and xCx_C be the number of points CC gets. Since xA=xB=xCx_A = x_B = x_C, to maximize xAx_A, we can just maximize xA+xB+xCx_A + x_B + x_C. But in each match, if one team wins then the total sum increases by 33 points, whereas if they tie, the total sum increases by 22 points. So, it is best if there are the fewest ties possible.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.