7. In a convex pentagon $A B C D E$, a point $M$ is taken on side $A E$, and a point $N$ is taken on side $D E$. Segments $C M$ and $B N$ intersect at point $P$. What is the smallest possible area of the pentagon $A B C D E$, given that the quadrilaterals $A B P M$ and $D C P N$ are parallelograms with areas param1 and param2, respectively?

| param1 | param2 | Answer |
| :---: | :---: | :---: |
| 8 | 9 | 29 |
| 10 | 45 | 85 |
| 9 | 50 | 89 |
| 6 | 75 | 111 |
| 8 | 49 | 85 |

## Condition

In a convex pentagon $A B C D E$, a point $M$ is taken on side $A E$ and a point $N$ is taken on side $D E$. Segments $C M$ and $B N$ intersect at point $P$. What is the smallest possible area of the pentagon $A B C D E$, given that the quadrilaterals $A B P M$ and $D C P N$ are parallelograms with areas 8 and 9, respectively?

## Answer.

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# Solution

Let the areas of parallelograms ABPM and DCPN, triangle BCP, and quadrilateral MPNE be $S_{1}, S_{2}, S_{3}$, and $S_{4}$, respectively. According to the problem, $A M \sqcup B \mathcal{B}$ and $P N \sqcup C \mathcal{D}$. Therefore, $\mathrm{AE} \sqcup \mathcal{B} N \cup C D$. Similarly, $D E \sqcup C M \sqcup A B$. Then, M PNE is also a parallelogram, and thus $S_{4}: S_{1}=N P: P B$ (parallelograms ABPM and M PNE have equal heights). Similarly, triangle $BCP$ and parallelogram $DCPN$ have equal heights, so $S_{3}: S_{2}=\frac{1}{2} B P: P N$. Multiplying the written equalities, we get: $\frac{S_{4}}{S_{1}} \cdot \frac{S_{3}}{S_{2}}=\frac{1}{2} \cdot$ Therefore, $S=S_{1}+S_{2}+$ $S_{3}+S_{4}=$

$=S_{1}+S_{2}+S_{3}+\frac{S_{1} S_{2}}{2 S_{3}} \geq S_{1}+S_{2}+2 \sqrt{S_{3} \cdot \frac{S_{1} S_{2}}{2 S_{3}}}=S_{1}+S_{2}+\sqrt{2 S_{1} S_{2}}=29$. The obtained minimum is achieved when $S_{3}=S_{4}$.