2006 Canadian MO Problems/Problem 4

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Problem

Consider a round robin tournament with $2n+1$ teams, where two teams play exactly one match and there are no ties. We say that the teams $X$, $Y$, and $Z$ form a cycle triplet if $X$ beats $Y$, $Y$ beats $Z$, and $Z$ beats $X$.

(a) Find the minimum number of cycle triplets possible.

(b) Find the maximum number of cycle triplets possible.

Solution

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See also

2006 Canadian MO (Problems)
Preceded by
Problem 3
1 2 3 4 5 Followed by
Problem 5

(a) Clearly the answer is 0.

(b) By using complementary counting, it is not hard to find that the answer is $\dfrac{n(n+1)(2n+1)}{6}$.