Points $K, L, M$ and $N$ are the midpoints of the sides $A B, B C, C D$ and $A D$ of parallelogram $A B C D$ with area $s$. Find the area of the quadrilateral formed by the intersections of the lines $A L, A M, C K$ and $C N$.

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Let the lines $A L$ and $C K$ intersect at point $P$, and the lines $A M$ and $C N$ intersect at point $Q$. Then $P$ and $Q$ are the points of intersection of the medians of triangles $A B C$ and $A D C$ respectively. Therefore,

$$
\begin{gathered}
S_{\triangle A P C}=\frac{1}{3} S_{\triangle A B C}=\frac{1}{3} \cdot \frac{1}{2} s=\frac{1}{6} s, S_{\triangle A Q C}=\frac{1}{3} S_{\triangle A D C}=\frac{1}{3} \cdot \frac{1}{2} s=\frac{1}{6} s . \\
\text { Hence, } \\
S_{A P C Q}=S_{\triangle A P C}+S_{\triangle A Q C}=\frac{1}{\overline{6}} s+\frac{1}{\overline{6}} s=\frac{1}{3} s .
\end{gathered}
$$

## Answer

$\frac{1}{3} s$.

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