2012 AMC 10B Problem 15

There are $\binom{6}{2} = 15$ games, and each produces exactly one win, so the six win totals add to $15$.

Could all six teams tie? Then each would have $15/6$ wins, not an integer. So at most $5$ teams can share the top spot.

Five is achievable. Call the teams $T_1, \dots, T_5$ and $L$. Let $L$ lose all five of its games, which gives each $T_i$ one win. The $T_i$ play $\binom52 = 10$ games among themselves; arrange them around a circle and let each team beat the next two teams clockwise and lose to the two preceding it. Each $T_i$ then wins $2$ and loses $2$ of those games. Every $T_i$ finishes with $1 + 2 = 3$ wins, $L$ finishes with $0$, and the five teams are tied for the most wins.

The answer is $\boxed{\textbf{(D)}\ 5}$.

Tournament "maximum tie" problems are two-step arguments: a counting bound (total wins is fixed and must divide evenly among tied teams) and an explicit schedule that meets the bound. The circular "beat the next $k$ teams" pattern is the standard way to make an odd number of teams all finish with equal records.

• Answering 6 without checking that 15 wins cannot be shared equally by 6 teams, or answering 3 or 4 for lack of an explicit construction.
• Trying to make the five tied teams win $3$ games each against one another (impossible, since only $10$ such games exist) and giving up on $5$.