Easy Chess Tournament Problem
- Mallika Prova
- Posts:6
- Joined:Thu Dec 05, 2013 7:44 pm
- Location:Mymensingh,Bangladesh
On a chess tournament 12 players took part, and any two of them played exactly one match.Any draw gives 0.5 point, win -1 point, loss -0.It turned out after the tournament that the first three together gained three times more points then the last five. How did the match between the seventh and eighth player end?
If you do what interests you , atleast one person is pleased.
- asif e elahi
- Posts:185
- Joined:Mon Aug 05, 2013 12:36 pm
- Location:Sylhet,Bangladesh
Re: Easy Chess Tournament Problem
Let $A,B,C$ be the set of first $3$,middle $4$ and last $5$ players respectively according to their rank and $U$ is the union set.Let $f(X,Y)$ denote the total score gained by the players of set $X$ against players of set $Y$.Easy to see that every match has total outcome of point $1$.
now $f(A,U)=f(A,A)+f(A,B\cup C) \leq \binom{3}{2} +3 \times 9=30$
and $f(C,U)=f(C,C)+f(C,A\cup B) \geq \binom{5}{2}+0=10$
So $f(A,U)\leq 3f(C,U)$.But we have them equal.
So we must have $f(C,U)=10$.Equality holds when $f(C,A\cup B)=0$.So players of $C$ loses all matches against the players of $A\cup B$.So the$8th$ player will lose against the $7th$ player.
now $f(A,U)=f(A,A)+f(A,B\cup C) \leq \binom{3}{2} +3 \times 9=30$
and $f(C,U)=f(C,C)+f(C,A\cup B) \geq \binom{5}{2}+0=10$
So $f(A,U)\leq 3f(C,U)$.But we have them equal.
So we must have $f(C,U)=10$.Equality holds when $f(C,A\cup B)=0$.So players of $C$ loses all matches against the players of $A\cup B$.So the$8th$ player will lose against the $7th$ player.