Problem 4. Given is a parallelogram $A B C D$. Let point $M$ be the midpoint of side $C D$, and point $N$ be the midpoint of side $A D$. The lines $A M$ and $B N$ intersect at point $P$. What fraction of the area of parallelogram $A B C D$ is the area of triangle $A N P$?
Solution. Let $S$ be a point on the line $BN$ such that $N$ is the midpoint of the segment $PS$. The segments $AD$ and $PS$ bisect each other, so quadrilateral $APDS$ is a parallelogram. Let $Q$ be the midpoint of side $AB$. From $\overline{AQ}=\overline{MC}$ and $AQ \| MC$

![](https://cdn.mathpix.com/cropped/2024_06_05_3a3538b5fdc92afac337g-14.jpg?height=308&width=492&top_left_y=478&top_left_x=987)
it follows that quadrilateral $AQCM$ is a parallelogram. Therefore, $CQ \| MA \| DS$. Furthermore, from $\overline{BQ}=\overline{QA}$ it follows that $\overline{BR}=\overline{RP}$, and from $\overline{CM}=\overline{MD}$ it follows that $\overline{RP}=\overline{PS}$. Therefore, $\overline{BR}=\overline{RP}=\overline{PS}$. But, $\overline{PS}=2 \overline{PN}$, so $\overline{BN}=\overline{BR}+\overline{RP}+\overline{PN}=2 \overline{PN}+2 \overline{PN}+\overline{PN}=5 \overline{PN}$. This means that $h_{B}=5 h_{P}$, where $h_{B}$ is the height of the parallelogram, and $h_{P}$ is the height in triangle $ANP$. Therefore,

$$
P_{ABCD}=\overline{AD} \cdot h_{B}=2 \overline{AN} \cdot 5 h_{P}=20 \cdot \frac{\overline{AN} \cdot h_{P}}{2}=20 P_{APN}
$$

i.e.

$$
P_{APN}=\frac{1}{20} P_{ABCD}
$$