6. Let's call "sawing" the following operation on a polygon (see fig.):

a) Each side of the polygon is divided into three equal parts.

b) The middle part is chosen as the base of an equilateral triangle located outside the polygon.

c) The base is removed and the other two sides are added.

Let $K_{0}$ be a square with side $2.5, K_{1}$ - the polygon obtained by sawing $K_{0}, K_{2}$ - obtained by sawing $K_{1}, \ldots, K_{2018}$ obtained by sawing $K_{2017}$ (see fig.). Find the area $S\left(K_{2018}\right)$. Round the answer to the hundredths according to standard mathematical rules.

![](https://cdn.mathpix.com/cropped/2024_05_06_b244a686143972116e53g-06.jpg?height=280&width=261&top_left_y=685&top_left_x=1046)

$\mathrm{K}_{0}$

![](https://cdn.mathpix.com/cropped/2024_05_06_b244a686143972116e53g-06.jpg?height=324&width=343&top_left_y=689&top_left_x=1259)

$\mathrm{K}_{1}$

![](https://cdn.mathpix.com/cropped/2024_05_06_b244a686143972116e53g-06.jpg?height=371&width=322&top_left_y=694&top_left_x=1575)

#
# Answer: 28.33.

Solution: On each step, the number of sides increases by a factor of 4. Therefore, on the $(n-1)$-th step, a $4^n$-gon is obtained. At the same time, the length of the side decreases by a factor of 3, so on the $(n-1)$-th step, the side length is $\frac{5}{2} \cdot 3^{1-n}$. Thus, on the $n$-th step, $4^n$ equilateral triangles are added, each with a side length of $\frac{5}{2} \cdot 3^{1-n}$. Therefore, the area increases on the $n$-th step by

$$
4^n \cdot \frac{\sqrt{3}}{4}\left(\frac{5}{2} \cdot 3^{1-n}\right)^2 = \frac{255 \sqrt{3}}{16} \cdot \left(\frac{4}{9}\right)^n
$$

The initial area $S_0 = \frac{25}{4}$, so $S_{2018} = \frac{25}{4} + \frac{255 \sqrt{3}}{16} \left(\left(\frac{4}{9}\right) + \left(\frac{4}{9}\right)^2 + \cdots + \left(\frac{4}{9}\right)^{2018}\right) \approx \frac{25}{4} + \frac{51 \sqrt{3}}{4} \approx 28.33$

## Olympiad for Schoolchildren "Conquer Vorobyovy Gory"

## Problems for 5-6 Grades

## Variant 1a (Kemerovo)