, 2025

Consider placing the robot’s movement onto the $xy$ plane.

Suppose that its starting point is $O(0,0)$, north is in the positive $y$ direction, and 1 unit in the plane
represents $1$ m travelled by the
robot.

The robot begins facing north and so in Step 1 it travels $2$ m to the point $A(0,2)$, as shown.

After turning 90$\degree$ left (to
face west) and moving $4$ m in the
direction that it is facing (Step 2 and Step 3), the robot is at $B(-4,2)$.

The robot then repeats these 3 steps, moving $2$ m forward to $C(-6,2)$, turning 90$\degree$ (to face south), and moving $4$ m in the direction that it is facing to
$D(-6,-2)$.

Repeating these 3 steps a 3rd time, the robot moves to $E(-6,-4)$ and then to $F(-2,-4)$.

Repeating these 3 steps a 4th time, the robot moves to $G(0,-4)$ and then to its starting point
$O(0,0)$, and is facing north.

[[IMAGE0]]

Therefore, each time the robot completes this sequence of 3 steps a
total of 4 times, it ends at $O(0,0)$
and is facing north.

Thus, if the robot completes this sequence of 3 steps $4\times6=24$ times, it ends at $(0,0)$ and is facing north.

Completing the sequence of 3 steps a 25th time, the robot moves to $B(-4,2)$ and is facing west (facing point
$C$).

Finally after completing the sequence a 26th time, the robot ends at
$D(-6,-2)$, and so its distance from
its starting point $O(0,0)$ is $x=\sqrt{(0-(-6))^2+(0-(-2))^2}=\sqrt{36+4}=\sqrt{40}$,
and so $x^2=40$.