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Difference between revisions of "2006 Romanian NMO Problems/Grade 9/Problem 4"
 
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==Problem==
 
==Problem==
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<math>\displaystyle 2n</math> students <math>\displaystyle (n \geq 5)</math> participated at table tennis contest, which took <math>\displaystyle 4</math> days. In every day, every student played a match. (It is possible that the same pair meets twice or more times, in different days) Prove that it is possible that the contest ends like this:
+
<math>2n</math> students <math>(n \geq 5)</math> participated at table tennis contest, which took <math>4</math> days. Every day, every student played a match. (It is possible that the same pair meets two or more times, in different days). Prove that it is possible that the contest ends like this:
  
 
* there is only one winner;
 
* there is only one winner;
  
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* there are <math>\displaystyle 3</math> students on the second place;
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* there are <math>3</math> students on the second place;
  
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* no student lost all <math>\displaystyle 4</math> matches.
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* no student lost all <math>4</math> matches.
  
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How many students won only a single match and how many won exactly <math>\displaystyle 2</math> matches? (In the above conditions)
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How many students won only a single match and how many won exactly <math>2</math> matches? (In the above conditions)
 
==Solution==
 
==Solution==
 +
Note that the 3 second place students obviously could not have only won one match, or won all 4 matches. I now claim that they could not have won exactly two matches, either.
 +
 +
Each day there were <math>n</math> matches, so at the end of the contest there were <math>4n</math> total points. Now if the three people in second place won exactly two matches, then <math>2n-4</math> people would have to had won exactly one match. The winner of the contest would have won at most 4 matches, so we have the inequality
 +
 +
<cmath>4+3\cdot 2 + (2n-4)\geq 4n</cmath>
 +
 +
Solving for <math>n</math> yields <math>n\leq 3</math>, which is clearly false. This is a contradiction in logic, so the three people in second place could not have won exactly two matches.
 +
 +
This shows that the three second-place finishers each won exactly three matches. Therefore the winner of the contest won all 4 matches. Now let <math>x</math> be the number of people who won two matches. It follows that <math>2n-x-4</math> people won one match. We now have the equation
 +
 +
<cmath>4 + 3\cdot 3 + 2x + 2n-x-4 = 4n</cmath>
 +
 +
Solving for <math>x</math> yields <math>x=2n-9</math>, so <math>\boxed{2n-9}</math> students won exactly two matches. It then follows that <math>\boxed{5}</math> people won a single match.
 +
 +
 
==See also==
 
==See also==
 
*[[2006 Romanian NMO Problems]]
 
*[[2006 Romanian NMO Problems]]
 
[[Category: Olympiad Combinatorics Problems]]
 
[[Category: Olympiad Combinatorics Problems]]

Latest revision as of 15:53, 11 December 2011

Problem

$2n$ students $(n \geq 5)$ participated at table tennis contest, which took $4$ days. Every day, every student played a match. (It is possible that the same pair meets two or more times, in different days). Prove that it is possible that the contest ends like this:

  • there is only one winner;
  • there are $3$ students on the second place;
  • no student lost all $4$ matches.

How many students won only a single match and how many won exactly $2$ matches? (In the above conditions)

Solution

Note that the 3 second place students obviously could not have only won one match, or won all 4 matches. I now claim that they could not have won exactly two matches, either.

Each day there were $n$ matches, so at the end of the contest there were $4n$ total points. Now if the three people in second place won exactly two matches, then $2n-4$ people would have to had won exactly one match. The winner of the contest would have won at most 4 matches, so we have the inequality

\[4+3\cdot 2 + (2n-4)\geq 4n\]

Solving for $n$ yields $n\leq 3$, which is clearly false. This is a contradiction in logic, so the three people in second place could not have won exactly two matches.

This shows that the three second-place finishers each won exactly three matches. Therefore the winner of the contest won all 4 matches. Now let $x$ be the number of people who won two matches. It follows that $2n-x-4$ people won one match. We now have the equation

\[4 + 3\cdot 3 + 2x + 2n-x-4 = 4n\]

Solving for $x$ yields $x=2n-9$, so $\boxed{2n-9}$ students won exactly two matches. It then follows that $\boxed{5}$ people won a single match.


See also

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