A negligibly small beetle starts from the origin of a plane rectangular coordinate system. First, it travels 1 unit to the right to the point $(1 ; 0)$. There, it makes a $90^{\circ}$ turn counterclockwise and travels $1 / 2$ unit, thus reaching the point ( $1 ; 1 / 2$ ). If it continues its journey in the same manner, that is, after each straight segment of its path, it turns $90^{\circ}$ counterclockwise and then travels half the distance it just covered, which point will the beetle approach more and more?
I. solution. In the steps of the ox, it alternately changes its first and second coordinates. With its odd-numbered steps, it modifies its first coordinate, always in the opposite direction and a quarter of the magnitude of the previous one. Therefore, the modification of the first coordinate at $k$, that is, after the $(2 k-1)$-th step, will be

$$
x_{2 k-1}=1-\frac{1}{4}+\ldots+\left(-\frac{1}{4}\right)^{k-1}=\frac{1-\left(-\frac{1}{4}\right)^{k}}{1-\left(-\frac{1}{4}\right)}
$$

and its value remains the same after the $2 k$-th step as well.

The second coordinate changes similarly, and its value after the $2 k$-th step is

$$
y_{2 k}=\frac{1}{2}-\frac{1}{8}+\ldots+\frac{1}{2}\left(-\frac{1}{4}\right)^{k-1}=\frac{1}{2} x_{2 k-1}
$$

and its value remains the same after the $(2 k+1)$-th step as well. Since the powers of 4 are different integers, the $k$-th one is greater than the $k$-th positive integer, that is, $4^{k}>k$. Therefore, $4^{k}$ grows beyond any bound as $k$ runs through the positive integers, and $4^{-k}$ approaches 0. Thus, the limit of $x_{n}$ is $4 / 5$, and the limit of $y_{n}$ is half of that, $2 / 5$.

Let $B_{n}$ denote the point $\left(x_{n} ; y_{n}\right)$ and $B$ the point $\left(\frac{4}{5} ; \frac{2}{5}\right)$. Since the distance between $B_{n}$ and $B$ is less than the sum $\left|x-x_{n}\right|+\left|y-y_{n}\right|$, and this sum approaches 0, the ox approaches the point $B\left(\frac{4}{5} ; \frac{2}{5}\right)$ more and more closely.

II. solution. Let $B_{n}$ still denote the position of the ox after the $n$-th step, and $B_{0}$ the origin.

![](https://cdn.mathpix.com/cropped/2024_05_02_396e1a29286e78feb885g-1.jpg?height=634&width=1066&top_left_y=1250&top_left_x=516)

The triangles $B_{n} B_{n+1} B_{n+2}$ for $n=0,1,2, \ldots$ are all similar, as they are right-angled and the ratio of the legs $B_{n} B_{n+1}$ and $B_{n+1} B_{n+2}$ is $2: 1$ regardless of $n$. Specifically, in the triangles $B_{0} B_{1} B_{2}$ and $B_{2} B_{3} B_{4}$, the angles at $B_{0}$ and $B_{2}$ are equal, and their sides are oppositely directed. Therefore, the hypotenuses of these triangles lie on the same line, and $B_{4}$ lies on the segment $B_{0} B_{2}$. Similarly, we find that $B_{5}$ lies on the segment $B_{1} B_{3}$, or the segment $B_{4} B_{5}$ can be obtained by a $16: 1$ reduction of the segment $B_{0} B_{1}$ from the point $B$. From this point onward, the entire path can be continued by applying this reduction to the broken line $B_{0} B_{1} B_{2} B_{3} B_{4}$, then to the result, and so on. In other words, if we apply a $16^{k}: 1$ reduction to the broken line $T_{0}=B_{0} B_{1} B_{2} B_{3} B_{4}$ from the center $B$, we get exactly the broken line $T_{k}=$ $B_{4 k} B_{4 k+1} B_{4 k+2} B_{4 k+3} B_{4 k+4}$. Since among the points of the broken line $T_{0}$, $B_{0}$ is the farthest from $B$, we get that the points of $T_{k}$ are at most $B_{0} B / 16^{k}$ away from $B$. Since this approaches 0 (which we can also verify as in the first solution, that $1 / 4^{k}$ approaches 0, although this can also be read directly from the latter statement), it follows that the ox gets closer and closer to the point $B$.

The equation of the line $B_{0} B_{2}$ is $y=x / 2$, and that of $B_{1} B_{3}$ is $y=2(1-x)$, so at their intersection, $x=4 / 5, y=2 / 5$, thus the coordinates of $B$ are $\left(\frac{4}{5} ; \frac{2}{5}\right)$.