4. In a football tournament, only teams from Small Town and Big City participated. There were 9 more teams from Big City than from Small Town. Each team met exactly once, with the winning team getting 1 point, the losing team 0 points, and no draws were possible. Teams from Big City scored 9 times as many points as teams from Small Town. Determine the maximum possible number of wins for the best team from Small Town.

The task should be solved independently. You have 210 minutes for solving. The use of notes, literature, or a pocket calculator is not allowed.

49th Mathematical Competition for High School Students in Slovenia

Velenje, April 16, 2005

## Tasks for 4th Year
III/4. Let the number of teams from Small Town be denoted by $x$. Then the number of teams from Big City is $x+9$. The teams from Small Town played $\frac{x(x-1)}{2}$ matches among themselves and thus scored $\frac{x(x-1)}{2}+k$ points, where $k$ is the number of wins over teams from Big City. The teams from Big City thus scored

$$
\frac{(x+8)(x+9)}{2}+x(x+9)-k
$$

points. It must hold that

$$
9\left(\frac{(x-1) x}{2}+k\right)=\frac{(x+8)(x+9)}{2}+x(x+9)-k
$$

thus $3 x^{2}-22 x+10 k-36=0$. Since $x$ is a natural number, the discriminant $4(229-30 k)$ must be a perfect square. Then $k \leq 7$ and thus $k=2$ or $k=6$. In the first case, $x=8$ and the best team from Small Town could have scored $7+2=9$ points. In the second case, $x=6$ and the best team from Small Town could have scored $5+6=11$ points.