A square frame is placed on a horizontal plane, touching one of its sides to a line e, on the same side of e. The lower left corner of each frame is hinged to the lower right corner of its left neighbor. Each side of a frame is as long as the diagonal of its right neighbor. The smallest frame is fixed, considered as the first one, and the rest, as a whole, are rotated counterclockwise by $45^{\circ}$. Then we fix the 2nd frame, and step by step, we proceed similarly with the rest, until a $45^{\circ}$ angle is formed at each hinge. In this final position, which numbered frames will have a common point with the line e?
We denote the consecutive frames as $K_{i}$, $i=0,1,2,3, \ldots, 49$, the lower right and left vertices as $A_{i}$ and $B_{i}$, respectively, and the other vertices in the order of traversal as $C_{i}, D_{i}$. Thus, $A_{i+1}$ coincides with $B_{i}$, and it is clear that the hinges lie on the line $e$ (1st figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_fb31f97172861a0d8db7g-1.jpg?height=678&width=1203&top_left_y=322&top_left_x=450)

After the first prescribed rotation - denoted by $F_{1}$ - the right side $A_{1} D_{1}$ of $K_{1}$ is perpendicular to the diagonal $B_{0} D_{0}$ of $K_{0}$, and thus its extension intersects the reflection of this diagonal over $e$, passing through the reflection of $D_{0}$ over $e$, which we denote as $O$ (2nd figure).

$O$ is simultaneously the reflection of the new position of $D_{1}$ over the line $B_{0} D_{0}=e_{1}$, on which the hinges now lie, since $A_{1} D_{1}=B_{0} D_{0}=B_{0} O=O A_{0} \cdot \sqrt{2}=O A_{1}$. Therefore, $O B_{1}=\sqrt{2} \cdot O B_{0}=(\sqrt{2})^{2} \cdot O A_{0}$. The post-$F_{1}$ positions of the frames $K_{1}, K_{2}, K_{3}, \ldots$ are obtained from the original positions of $K_{0}, K_{1}, K_{2}, \ldots$ by a rotation and dilation centered at $O$, with an angle of $+45^{\circ}$ and a dilation factor of $\sqrt{2}$ (of course, the image of $K_{49}$ constructed this way is not to be considered). We also note that after $F_{1}$, only $K_{0}$ is in the $45^{\circ}$ sector $D_{0} O B_{0}$, and that the diagonal $A_{1} C_{1}$ of $K_{1}$ lies on $e$ and remains there, as $K_{1}$ is fixed in this position.

Draw the rays $f_{j}$ from $O$ $(j=1,2 \ldots, 7)$, which form angles of $j \cdot 45^{\circ}$ with $O A_{0}=f_{0}$, i.e., $f_{1}$ coincides with $O A_{1}$, and denote the sector between $f_{j-1}$ and $f_{j}$ as $T_{j-1}$.

The second rotation $F_{2}$ occurs around $B_{1}=A_{2}$, and the set of frames $K_{2}, K_{3}, \ldots, K_{49}$ participates in it. Only $K_{1}$ remains in $T_{1}$, $B_{2} A_{2} \perp f_{2}$, and since $B_{2} A_{2}=B_{1} D_{1}=B_{1} O$, $B_{2}$ lands on $f_{2}$, $O B_{2}=\sqrt{2} \cdot O B_{1}=(\sqrt{2})^{3} \cdot O A_{0}$, and $K_{2}$ is fixed here.

By alternately performing the rotations $F_{3}, F_{4}, \ldots$ and fixing $K_{3}, K_{4}, \ldots$, we similarly obtain that each frame $K_{i}$ is fixed in the sector $T_{j}$ where the index $j$ is the smallest non-negative remainder of $i$ divided by 8. The fixed positions of $B_{i}$ and $A_{i}$ are on the boundaries of $T_{j}$, on $f_{j-i}$ and $f_{j}$, respectively, such that $O B_{i}=(\sqrt{2})^{i} \cdot O A_{0}$ and $O A_{i}=(\sqrt{2})^{i-1} \cdot O A_{0}$.

One ray of $e$ falls into the sectors $T_{1}$ and $T_{6}$, so $e$ will have a common point with every frame $K_{i}$ whose smallest non-negative remainder of $i$ divided by 8 is 1 or 6. However, the segments of $e$ falling into $T_{0}$ and $T_{7}$ are already too short for the smallest applicable frames $K_{8}$ and $K_{7}$ to fit in the described positions, so no other $K_{i}$ will have a common point with $e$ except for the ones mentioned. Therefore, the frames that have a common point with $e$ are:

$$
\begin{gathered}
0, \quad 1, \quad 9, \quad 17, \quad 25, \quad 33, \quad 41, \quad 49, \\
6, \quad 14, \quad 22, \quad 30, \quad 38, \quad 46
\end{gathered}
$$

Péter Krasznai (Győr, Czuczor G. Bencés Gymnasium, 3rd grade)