A regular triangle is rotated in a positive direction around its center, first by $3^{\circ}$, then by $9^{\circ}$, and then by $27^{\circ}$, and in the $k$-th step by $3^{k}$ degrees. How many different positions can the triangle take as a result of these rotations?
The first three transformations result in a $3^{\circ}$, a $3+9=12^{\circ}$, and a $3+9+27=39^{\circ}$ rotation, respectively; these place the equilateral triangle in different positions, distinct from the initial position. However, the result of four consecutive transformations (relative to the position after the $k$-th rotation) is

$$
3^{k+1}+3^{k+2}+3^{k+3}+3^{k+4}=3^{k}(3+9+27+81)=3^{k} \cdot 120
$$

degrees of rotation, which, being a multiple of $120^{\circ}$, does not change the position of the equilateral triangle. Therefore, the position after the $(k+4)$-th rotation is the same as after the $k$-th rotation (for all $k$), meaning the triangle will assume a total of 4 different positions during the sequence of movements.

Koch Dénes (Linz, Akad. Gymn., 11. o.t.)

Note. Many noticed that the fourth rotation maps the triangle back to itself, but did not prove that only these four positions can occur subsequently. They received 1 point for their solution.