2016 AMC 10B Problems/Problem 22
Contents
Problem
A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won games and lost games; there were no ties. How many sets of three teams were there in which beat , beat , and beat
Solution 1
There are teams. Any of the sets of three teams must either be a fork (in which one team beat both the others) or a cycle:
But we know that every team beat exactly other teams, so for each possible at the head of a fork, there are always exactly choices for and as beat exactly 10 teams and we are choosing 2 of them. Therefore there are forks, and all the rest must be cycles.
Thus the answer is which is .
Solution 2 (Cheap Solution)
Since there are teams and for each set of three teams there is a cycle, there are a total of cycles of three teams. Because about of the cycles satisfy the conditions of the problems, our answer is close to . Looking at the answer choices, we find that is closer to than any other answer choices, and that the next closest is which is twice of , so our answer is which is .
Speaking of cheap solutions, you can let the teams be and have beat and lose to You can choose any random team as for ways but overcount since set, so divide by so multiply by Now we can proceed w/casework (which isn't too hard). Eventually you realize the sum is .
Solution 3 (Circle)
Let's arrange all the teams in a circle and assume that each team won against the first 10 teams clockwise of themselves, and lost against the first 10 teams counter-clockwise of themselves.
Consider a working set of teams: moving clockwise, we know that the first team beat the team clockwise of itself, that team beat the next team clockwise of itself, and the final team beat the first team, which would be clockwise of itself. However, we also must remember that if the first team beat the second team, moving clockwise, the second team cannot be more than teams away from the first team; the same applies to the second and third team, and the third and first team.
Let's say, WLOG, that the first team, team , is at position out of on the circle.
If team is to beat team , since we are assuming each team beats the 10 members clockwise of themselves, there are places on the circle that team could be: positions through . Also, if team is to beat team , team must be located from positions to .
If Team is in position , must be located from position to position , since beats . We also just found that 's position must be between and inclusive. So, when is in position , can only be in position . When is in position , can be in position or . In general, when is in position , there are choices for where can be. So, there are ways to place and . There are players to choose as player , and each working set will be counted times, so our final answer is .
Solution 4 (Aggregate Counting)
This is a Graph Theory problem with directed graph. There are teams in total. WLOG, pick team , there are teams that lost to and teams that won over . Call the group of teams that lost to group , and the group of teams that won over group .
Any team from group that won a team from group will form a cycle with . Now we need to count how many teams in group won over a team from group . The total number of wins in group is . There are wins among the teams inside group . So group has wins over group .
See Also
2016 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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All AMC 10 Problems and Solutions |
2016 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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