2006 Romanian NMO Problems/Grade 9/Problem 4

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.

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2006 Romanian NMO Problems/Grade 9/Problem 4

Problem

Solution

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