Task B-4.3. (20 points) Each side of a square, of length $a$, is divided in the ratio $1: x(x>1)$. The obtained points are the vertices of a new square inscribed in the original one. The same procedure is repeated with each new square. What is the sum of the areas of the original and all thus obtained squares?
Solution. The area of the original square is $P_{1}=a^{2}$;

Let $|A H|=|E B|=k$. Then $|A E|=|B F|=k x$. Since $|A E|+|E B|=a$, we get $k x+k=a$ i.e. $k=\frac{a}{x+1}$. The side $a_{2}=|E H|$ is found using the Pythagorean theorem, i.e.

$$
a_{2}^{2}=|A E|^{2}+|A H|^{2}=(k x)^{2}+k^{2}=k^{2}\left(x^{2}+1\right)=\frac{a^{2}\left(x^{2}+1\right)}{(x+1)^{2}}
$$

The area of the inscribed square is $P_{2}=a_{2}^{2}=\frac{a^{2}\left(x^{2}+1\right)}{(x+1)^{2}}$;

The area of the new inscribed square is $P_{3}=a_{3}^{2}=\frac{a^{2}\left(x^{2}+1\right)}{(x+1)^{2}} \cdot \frac{\left(x^{2}+1\right)}{(x+1)^{2}}=\frac{a^{2}\left(x^{2}+1\right)^{2}}{(x+1)^{4}}$.

The process continues analogously.

(8 points)

The sum of the areas of all squares is

![](https://cdn.mathpix.com/cropped/2024_05_30_9c56249806d755c5106ag-15.jpg?height=414&width=420&top_left_y=221&top_left_x=824)

$$
P=a^{2}+\frac{a^{2}\left(x^{2}+1\right)}{(x+1)^{2}}+\frac{a^{2}\left(x^{2}+1\right)^{2}}{(x+1)^{4}}+\frac{a^{2}\left(x^{2}+1\right)^{3}}{(x+1)^{6}}+\ldots
$$

(4 points)

The desired sum of areas is the sum of a geometric series with the quotient $q=\frac{x^{2}+1}{(x+1)^{2}}=\frac{x^{2}+1}{x^{2}+2 x+1}<1$, which means the series converges.

Since the first term is equal to $a^{2}$, the desired area is

$$
P=\frac{a^{2}}{1-q}=\frac{a^{2}}{1-\frac{x^{2}+1}{(x+1)^{2}}}=a^{2} \cdot \frac{(x+1)^{2}}{2 x}
$$

(5 points)