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

Revision as of 01:15, 28 October 2022

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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 ==Solution==
 {{solution}}
-(a) The answer is 0 cycle triplets. There is no possible value less than it, and it can be achieved if team <math>T_i</math> beats team <math>T_j</math> for all <math>i>j</math>.
+[[Image:CMO2006Question4.jpg |600px|thumb|left|]]
-(b) Every triplet of teams is either a cycle triplet or a "dominant" triplet(One team defeats the two others). There are a total of <math>\dbinom{2n+1}{3}</math> triplets, so we count the minimum number of "dominant" triplets.
-Note that a "dominant" triplet is uniquely determined by the dominant team. Let team <math>T_i</math> have <math>a_i</math> wins in the tournament. Then choosing any pair of those teams that <math>T_i</math> beat and joining them with <math>T_i</math> gives exactly one "dominant" triplet. Thus, the number of dominant triplets is <math>\sum_{i=1}^{n} \dbinom{a_i}{2}</math>.
-But since exactly <math>\dbinom{2n+1}{2}</math> games were played, <math>\sum_{i=1}^{n} a_i = \dbinom{2n+1}{2}</math>.
-<math>\sum_{i=1}^{n} \dbinom{a_i}{2}=\frac{\sum_{i=1}^{n} {a_{i}^{2}} - \sum_{i=1}^{n} a_i}{2}=\frac{\sum_{i=1}^{n} {a_{i}^{2}} - \dbinom{2n+1}{2}}{2}</math>.
 ==See also==

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

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

Revision as of 01:15, 28 October 2022

Problem

Solution

See also