## Task B-4.6.

A square $K_{1}$ with an area of 1 is circumscribed by a circle, and to this circle, a square $K_{2}$ is circumscribed, whose sides are parallel to the sides of the initial square $K_{1}$. The area between the squares $K_{1}$ and $K_{2}$ is colored gray. Then, a circle is circumscribed around the square $K_{2}$, and to this circle, another square $K_{3}$ is circumscribed in the same manner. The area between the squares $K_{2}$ and $K_{3}$ is left uncolored. This process is repeated until there are 2024 gray-colored areas and 2025 uncolored areas ( $K_{1}$ is counted among the uncolored areas). What is the area of all the gray-colored areas, and what is the area of all the uncolored areas?
## Solution.

The first few steps in coloring the area between two squares are shown in the figure:

![](https://cdn.mathpix.com/cropped/2024_05_30_803c5e1b47d2c202a3a9g-30.jpg?height=412&width=414&top_left_y=368&top_left_x=752)

There are a total of $2024+2025=4049$ areas that are either colored or uncolored, and the number of constructed squares is 4049.

Let $a_{1}, a_{2}, \ldots, a_{4049}$ be the side lengths of the squares $K_{1}, K_{2}, \ldots, K_{4049}$, and $P_{1}, P_{2}, \ldots, P_{4049}$ their areas. The radii of the circles circumscribed around the squares will be denoted as $r_{1}, r_{2}, \ldots, r_{4048}$.

The area of the colored region is equal to the sum $P_{2}-P_{1}+P_{4}-P_{3}+\cdots+P_{4048}-P_{4047}$, and the area of the uncolored region is the sum $P_{1}+P_{3}-P_{2}+P_{5}-P_{4}+\cdots+P_{4049}-P_{4048}$.

The side lengths of the squares are, in order:

$$
\begin{gathered}
a_{1}=1 \\
a_{2}=2 r_{1}=a_{1} \sqrt{2}=\sqrt{2} \\
a_{3}=2 r_{2}=a_{2} \sqrt{2}=2 \\
a_{4}=2 r_{3}=a_{3} \sqrt{2}=2 \sqrt{2}
\end{gathered}
$$

The side lengths of the squares form a geometric sequence with a common ratio $q=\sqrt{2}$ and the first term $a_{1}=1$, so the general term is

$$
a_{n}=(\sqrt{2})^{n-1}=2^{\frac{n-1}{2}}
$$

The area of the square $K_{n}$ is $P_{n}=\left(a_{n}\right)^{2}=2^{n-1}$. Clearly, the areas of the squares form a geometric sequence with a common ratio $q_{P}=2$.

The area of the colored region $P_{O}$ is

$P_{O}=P_{2}-P_{1}+P_{4}-P_{3}+\cdots+P_{4048}-P_{4047}=-P_{1}+P_{2}-P_{3}+P_{4}+\cdots-P_{4047}+P_{4048}$.

Notice that the sum of the colored areas is equal to the sum of 4048 terms of a geometric sequence with a common ratio $q_{O}=-q_{P}=-2$ and the first term $-P_{1}=-1$. Therefore,

$$
P_{O}=-1 \cdot \frac{(-2)^{4048}-1}{-2-1}=\frac{1}{3}\left(2^{4048}-1\right)
$$

The area of the uncolored region $P_{N}$ is

$P_{N}=P_{1}+P_{3}-P_{2}+P_{5}-P_{4}+\cdots+P_{4049}-P_{4048}=P_{1}-P_{2}+P_{3}-P_{4}+\cdots-P_{4048}+P_{4049}$.

As with the sum of the colored areas, we conclude that the sum of the uncolored areas is equal to the sum of 4049 terms of a geometric sequence with a common ratio $q_{N}=-q_{P}=-2$ and the first term $P_{1}=1$. Therefore,

$$
P_{N}=1 \cdot \frac{(-2)^{4049}-1}{-2-1}=-\frac{1}{3}\left(-2^{4049}-1\right)=\frac{1}{3}\left(2^{4049}+1\right)
$$

Note: Students receive 1 point for determining the total number of constructed squares, 1 point each for determining the sum of the colored and uncolored areas, or which is the last colored and uncolored area in the desired sum. Three points for recognizing the geometric sequences (side lengths and areas), stating the common ratio and the general term of these sequences. Finally, 2 points each for calculating the total colored and uncolored area.

Students can also calculate the sum of the colored and uncolored areas in other ways, for example, they can calculate the area by regions:

$$
\begin{gathered}
P_{0}=\left(P_{2}-P_{1}\right)+\left(P_{4}-P_{3}\right)+\cdots+\left(P_{4048}-P_{4047}\right)=1+4+16+\cdots+2^{4046}= \\
1 \cdot \frac{4^{2024}-1}{4-1}=\frac{1}{3}\left(2^{4048}-1\right) \\
P_{N}=P_{1}+\left(P_{3}-P_{2}\right)+\left(P_{5}-P_{4}\right)+\cdots+\left(P_{4049}-P_{4048}\right)=1+\left(2+8+16+\cdots+2^{4046}\right)= \\
1+2 \cdot \frac{4^{2024}-1}{4-1}=1+\frac{2}{3}\left(2^{4048}-1\right)=\frac{1}{3}\left(2^{4049}+1\right)
\end{gathered}
$$

Additionally, a student can conclude that the total area of the uncolored parts is equal to half the total area of the colored parts plus the area of the initial square, and thus calculate the uncolored area. The same applies if the colored area is calculated as twice the uncolored area (without the initial square). In each of these cases, the student receives all the points allocated for calculating the areas.