A cat always jumps in leaps of length $1 \mathrm{~m}$. Initially, this cat is at point A in the figure below, which is $2 \mathrm{~m}$ away from point $O$. Then, she jumps to point $B$, which is $1 \mathrm{~m}$ away from point $A$ and such that the segment $A B$ is perpendicular to the segment $O A$. Next, the cat jumps from point $B$ to point $C$, which is $1 \mathrm{~m}$ away from point $B$ and such that $B C$ is perpendicular to the segment $O B$, and so on.

![](https://cdn.mathpix.com/cropped/2024_05_01_d1e5bbfbdffc7558a598g-06.jpg?height=672&width=651&top_left_y=708&top_left_x=584)

a) What is the length of the segment $O B$?

b) What is the length of the segment $O C$?

c) After 2014 jumps, how far from point $O$ will the cat be? After how many jumps will she be exactly $45 \mathrm{~m}$ from point $O$?
Solution

In this solution, all distances are given in meters.

a) Observe the figure below:

![](https://cdn.mathpix.com/cropped/2024_05_01_d1e5bbfbdffc7558a598g-25.jpg?height=299&width=579&top_left_y=469&top_left_x=615)

By the Pythagorean Theorem, we have:

$$
x^{2}=2^{2}+1^{2}
$$

Thus, we have $x=\sqrt{5}$, which is the length of the segment $O B$.

b) Observe the figure below:

![](https://cdn.mathpix.com/cropped/2024_05_01_d1e5bbfbdffc7558a598g-25.jpg?height=494&width=579&top_left_y=1170&top_left_x=615)

Again, by the Pythagorean Theorem, we have:

$$
y^{2}=(\sqrt{5})^{2}+1^{2}
$$

From which we conclude that $y=\sqrt{6}$. Therefore, the length of the segment $B C$ is equal to $\sqrt{6}$.

c) Repeating the previous process, always using the Pythagorean Theorem, we obtain $\sqrt{7}, \sqrt{8}, \sqrt{9}$, and so on. Observe:

|  | Distance |
| :---: | :---: |
| $1^{\text{st}}$ jump | $\sqrt{5}$ |
| $2^{\text{nd}}$ jump | $\sqrt{6}$ |
| $3^{\text{rd}}$ jump | $\sqrt{7}$ |
| $\vdots$ | $\vdots$ |
| $2014^{\text{th}}$ jump | $x$ |

Thus, we deduce that $x=\sqrt{2018}$. Therefore, after 2014 jumps, the cat will be at a distance of $\sqrt{2018} \mathrm{~m}$ from point $O$. To find out after how many jumps the cat will be exactly $45 \mathrm{~m}$ from point $O$, we need to find a number such that the square root of this number is equal to 45. That is, this number must be equal to $45^{2}=2025$. Looking at the table above, we can deduce that the cat will be at a distance of $\sqrt{2025} \mathrm{~m}$ from point $O$ after 2021 jumps.