Given a rectangle $A B C D$. Through point $A$ we draw a line that intersects segment $C D$ at point $X$ such that the areas of the resulting figures satisfy $S_{A X D}: S_{A B C X}=1: 2$. Through point $X$ we draw a line that intersects segment $A B$ at point $Y$ such that $S_{A X Y}: S_{Y B C X}=1: 2$. Finally, through point $Y$ we draw a line that intersects segment $X C$ at point $Z$ such that $S_{X Y Z}: S_{Y B C Z}=$ $=1: 2$.

Calculate the ratio of the areas $S_{A X D}: S_{A X Z Y}$.
From the given, $S_{X Y Z}: S_{Y B C Z}=1: 2$. If we denote the area of triangle $X Y Z$ as $S$, the area of quadrilateral $Y B C Z$ is $2 S$, see the figure.

![](https://cdn.mathpix.com/cropped/2024_04_17_908afcea70d98edd1123g-5.jpg?height=576&width=811&top_left_y=797&top_left_x=654)

Furthermore, it is given that $S_{A X Y}: S_{Y B C X}=1: 2$ and from the previous, the area of quadrilateral $Y B C X$ is $2 S + S = 3 S$. Therefore, the area of triangle $A X Y$ must be $3 S / 2 = 1.5 S$. It is also given that $S_{A X D}: S_{A B C X}=1: 2$. The area of quadrilateral $A B C X$ is $2 S + S + 1.5 S = 4.5 S$. Thus, the area of triangle $A X D$ is $4.5 S / 2 = 2.25 S$. The desired ratio $S_{A X D}: S_{A X Z Y}$ is $2.25 S: (1.5 S + S)$, i.e., $2.25: 2.5$. After expanding by four, we get the ratio $9: 10$.

Note. The problem can also be solved by determining the lengths of segments $|D X|, |A Y|$, and $|X Z|$ in terms of the length of segment $|A B|$, expressing the areas of the required figures, and then finding their ratio: If we denote the length of segment $|A B|$ as $a$, then from the requirement $S_{A X D}: S_{A B C X}=1: 2$, it is easily derived that $|D X|=\frac{2}{3} a$. Similarly, from the remaining two equalities, we get $|A Y|=\frac{4}{9} a$ and $|X Z|=\frac{8}{27} a$. The ratio $S_{A X D}: S_{A X Z Y}$ is then equal to $\frac{|D X| \cdot|B C|}{2}: \frac{(|A Y|+|X Z|) \cdot|B C|}{2}=\frac{|D X|}{|A Y|+|X Z|}$. After substitution and simplification, it results in $9: 10$.