The midpoint of each side of the parallelogram is connected to the ends of the opposite side. Find the area of the octagon formed by the intersections of the drawn segments, if the area of the parallelogram is 1.

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Let $K, L, M$ and $N$ be the midpoints of the sides $AB, BC, CD$ and $AD$ of parallelogram $ABCD$ with area $1$, and $O$ be the center of the parallelogram. Then the area of parallelogram $KBL O$ is $\frac{1}{t}$, and the area of triangle $KLO$ is $\frac{1}{8}$.

Let the lines $AL$ and $BN$ intersect at point $P$, the lines $BM$ and $CK$ at point $Q$, and the lines $AL$ and $CK$ at point $H$. Then $P, H$ and $Q$ are three consecutive vertices of the octagon mentioned in the problem. Here, $P$ is the midpoint of side $KO$ of triangle $KLO$, $Q$ is the midpoint of side $LO$, and $H$ is the intersection of the medians $LP$ and $KQ$ of this triangle, so the area of quadrilateral $OPHQ$ is one-third of the area of triangle $KLO$, i.e., $\frac{1}{3} \cdot \frac{1}{8}=\frac{1}{2 t}$. It remains to note that the area of the octagon is four times the area of quadrilateral $OPHQ$, i.e., $4 \cdot \frac{1}{2 t}=\frac{1}{6}$.

## Answer

$\frac{1}{6}$.