Given a quadrilateral $A B C D$ with an area of 1. From its internal point O, perpendiculars OK, OL, OM, and ON are dropped to the sides $A B, B C, C D$, and $D A$ respectively. It is known that $A K \geq K B, B L \geq L C, C M \geq M D$, and $D N \geq N A$. Find the area of the quadrilateral $K L M N$.
From two inclined lines drawn from one point, the one with the larger projection is larger. Therefore, from the inequalities given in the problem, it follows that $O A \geq O B \geq O C \geq O D \geq O A$. Hence, $O A=O B=O C=O D$, which means $O$ is the center of the circle circumscribed around quadrilateral $A B C D$ (see figure). Therefore, points $K, L, M$, and $N$ are the midpoints of the sides of $A B C D$. Consequently, according to problem $\underline{56493}$ a) $S_{K L M N}=1 / 2 S_{A B C D}=1 / 2$.

![](https://cdn.mathpix.com/cropped/2024_05_06_ccb5665c9ecd710ea159g-06.jpg?height=718&width=755&top_left_y=470&top_left_x=663)

## Answer

0.5 .

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