4. Variant 1. An ant, starting from point A, goes $1+\frac{1}{10}$ cm north, then $2+\frac{2}{10}$ cm west, then $3+\frac{3}{10}$ cm south, then $4+\frac{4}{10}$ cm east, then $5+\frac{5}{10}$ cm north, then $6+\frac{6}{10}$ cm west, and so on. After 1000 steps, the ant is at point B. Find the distance between points A and B in centimeters in a straight line. In the answer, write the square of the found distance.
Answer: 605000.

Solution. Let's divide 1000 steps into quartets. After each quartet, the ant will move southeast relative to its current position, by a distance equal to the diagonal of a square with side 2.2, i.e., $\sqrt{2.2^{2}+2.2^{2}}=\sqrt{9.68}$. After 250 such quartets, the ant will be at a distance of $250 \cdot \sqrt{9.68}$ from point A. The square of this distance is $250^{2} \cdot 9.68=605000$.