2012 AMC 10B Problem 15

This asks for a maximum, so there are two parts: show that $6$ tied teams is impossible, then show that $5$ tied teams can really happen.

Part 1: six teams cannot all tie.

1. Count the games. Each of the $6$ teams plays the other $5$, which is $6 \times 5 = 30$, but that counts every game twice (once for each team). So there are $15$ games.
2. Each game gives exactly one win, so the six teams have $15$ wins altogether.
3. If all six tied, each would have $15 \div 6 = 2.5$ wins. That is not a whole number, so it can't happen.

Part 2: five teams can tie.

4. Call the teams $A, B, C, D, E$ and $F$. Let $F$ lose every game. That gives each of $A$ through $E$ one win.
5. Now the five teams $A$ to $E$ play $10$ games among themselves. We want each of them to win exactly $2$ of their $4$ games. Seat them in a circle $A, B, C, D, E$ and let each team beat the next two teams around the circle:
- $A$ beats $B$ and $C$;
- $B$ beats $C$ and $D$;
- $C$ beats $D$ and $E$;
- $D$ beats $E$ and $A$;
- $E$ beats $A$ and $B$.
6. Check that every game is listed exactly once: this list has $10$ results, and each pair of teams appears once (for example $A$ vs $D$ appears only as "$D$ beats $A$").
7. So each of $A$ through $E$ has $1 + 2 = 3$ wins, and $F$ has $0$. Five teams are tied for the most wins.

Six is impossible and five is possible, so the maximum is $5$.

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

For "most teams tied" questions, the total number of wins is fixed, and it has to divide evenly among the tied teams. That gives the upper limit. Then you need an actual schedule to prove the limit can be reached. The circle trick, where each team beats the next few teams around, is a standard way to give an odd number of teams equal records.

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 $\boxed{\textbf{(D)}\ 5}$.