Difference between revisions of "2006 Canadian MO Problems/Problem 4"

 
(4 intermediate revisions by 2 users not shown)
Line 6: Line 6:
 
(b) Find the maximum number of cycle triplets possible.
 
(b) Find the maximum number of cycle triplets possible.
 
==Solution==
 
==Solution==
{{solution}}
+
 
 +
 
 +
[[Image:CMO2006Question4.jpg |700px]]
 +
 
  
 
==See also==
 
==See also==
Line 12: Line 15:
  
 
{{CanadaMO box|year=2006|num-b=3|num-a=5}}
 
{{CanadaMO box|year=2006|num-b=3|num-a=5}}
 
(a) Clearly the answer is 0.
 
 
(b) By using complementary counting, it is not hard to find that the answer is <math>\dfrac{n(n+1)(2n+1)}{6}</math>.
 

Latest revision as of 17:20, 28 November 2023

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

CMO2006Question4.jpg


See also

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