8. (4 points) Given a triangle $ABC$ with sides $AB=13$, $BC=14$, $AC=15$. On side $AB$, point $K$ is marked; on side $BC$, point $L$ is marked; on side $AC$, point $N$ is marked. It is known that $BK=\frac{14}{13}$, $AN=10$, $BL=1$. A line parallel to $KL$ is drawn through point $N$, intersecting side $BC$ at point $M$. Find the area of quadrilateral $KLMN$.

Answer: $\frac{36503}{1183}=30 \frac{1013}{1183}$.
Solution:

The area of triangle $ABC$ by Heron's formula is 84.

$$
\begin{gathered}
S_{KBL}=\frac{KB \cdot BL}{AB \cdot BC} \cdot S_{ABC}=\frac{\frac{14}{13} \cdot 1}{13 \cdot 14} \cdot 84=\frac{84}{169} \\
S_{AKN}=\frac{AK \cdot AN}{AB \cdot AC} \cdot S_{ABC}=\frac{\left(13-\frac{14}{13}\right) \cdot 10}{13 \cdot 15} \cdot 84=\frac{8680}{169}
\end{gathered}
$$

Draw a line through point $A$ parallel to $KL$. It will intersect side $BC$ at point $X$. Triangles $BKL$ and $BAX$ are similar, so $BX=\frac{AB \cdot BL}{BK}=\frac{\frac{13 \cdot 1}{11}}{\frac{14}{13}}=\frac{169}{14}$, hence $CX=BC-BX=14-\frac{169}{14}=\frac{27}{14}$.

Triangle $MNC$ is similar to triangle $XAC$ with a similarity ratio of $\frac{1}{3}$.

$$
\begin{gathered}
S_{MNC}=\frac{1}{9} S_{XAC}=\frac{1}{9} \cdot \frac{XC}{BC} \cdot S_{ABC}=\frac{1}{9} \cdot \frac{\frac{27}{14}}{14} \cdot 84=\frac{9}{7} \\
S_{KLMN}=S_{ABC}-S_{KBL}-S_{ANK}-S_{MNC}=84-\frac{84}{169}-\frac{8680}{169}-\frac{9}{7}=\frac{36503}{1183}=30 \frac{1013}{1183}
\end{gathered}
$$

## 9th Grade  2nd Variant

1. (2 points) In an infinite volleyball tournament, 33 teams are participating. Each morning, the organizer schedules the matches for the day. Each day, 32 teams are paired to play one match each, and one team rests. Each match ends with one team winning, there are no draws.

Prove that the organizer can schedule the matches in such a way that eventually, there will be a team that loses six consecutive matches.