$8 \mathrm{db} r$ radius circular disks are arranged such that 7 of them are fixed on the table with their centers at the vertices and the center of a $2 r$ side regular hexagon, while the 8th disk touches one of the 6 outer disks. The 8th disk is rolled around the others without slipping until it returns to its initial position. How many rotations does the disk make around its own center during this process? (Explain your answer.)
In the given situation, the annular disks centered at the adjacent vertices of the hexagon, $k_{1}, k_{2}, k_{3}, k_{4}, k_{5}, k_{6}$, touch each other, and each of these disks touches the annular disk centered at the hexagon's center, $k_{7}$. The moving disk $k_{8}$ traces a closed path, so we can describe its movement starting from any arbitrary position.

![](https://cdn.mathpix.com/cropped/2024_05_02_6ac531c0e3aaefcfde26g-1.jpg?height=727&width=702&top_left_y=306&top_left_x=706)

Let's consider an initial position where $k_{8}$ touches two stationary (outer) disks, $k_{1}$ and $k_{2}$, whose centers are $O_{1}$ and $O_{2}$, respectively. Let the position of the center of $k_{8}$ be $M_{1}$, and the points of contact be $T_{1}$ and $T_{2}$. The 7 fixed disks have a 6-fold rotational symmetry, so it is sufficient to determine how much a selected radius of $k_{8}$ rotates around its own center $M$ until $k_{8}$ simultaneously touches $k_{2}$ and the other adjacent outer disk, the disk $k_{3}$ centered at $O_{3}$. During a complete traversal, this rotation will be six times as much. In the new position, let the center of $k_{8}$ be $M_{2}$, and the points of contact be $U_{2}$ and $U_{3}$. As $k_{8}$ rolls from $M_{1}$ to $M_{2}$, it continuously touches $k_{2}$ along the arc $T_{2} U_{2}$. This arc is one-third of a circle, because six circles of equal radius can be placed around a circle such that each touches the original circle and two others (as with the first six stationary disks $k_{7}$). Out of the 6 circles that touch $k_{2}$, 3 are drawn, and the intermediate positions of $k_{8}$ between $M_{1}$ and $M_{2}$ are the other 3, with the points of contact with $k_{2}$ being one-third of the way between the contact points of the outermost circles.

Considering $k_{2}$ as a conventional clock face, we can say that $k_{8}$ rolls along the arc from 12 o'clock to 4 o'clock. Let's also consider $k_{8}$ as a clock face. If in the initial position it was aligned with $k_{2}$, then the 6 o'clock mark fell at $T_{2}$, so it is useful to observe the change in the direction of the $M 6$ radius during the movement.

At any instant of rolling, let the point $V$ of $k_{8}$ coincide with the point $T$ of $k_{2}$. Since there is no slipping, the arc (12) $T$ is equal to the arc $6^{\prime} V$, and from the equality of the radii, the angles $6^{\prime} M^{\prime} V$ and (12) $O_{2} T$ are equal, and thus the $M^{\prime} 6$ radius has rotated twice the angle that the line $O_{2} M^{\prime}$ of the pair of circles $k_{2}, k_{8}^{\prime}$ has rotated relative to $O_{2} M_{1}$. Therefore, by the time the center of $k_{8}$ reaches $M_{2}$, the disk itself has rotated $2 \cdot 120^{\circ}=240^{\circ}$, or $2/3$ of a full rotation. During the complete traversal, this will be six times as much, so $k_{8}$ makes 4 full rotations around its own center.

Tamás Endre (Budapest, I. István g. II. o. t.)[^0]


[^0]:    ${ }^{1} \mathrm{The}$ diagram shows $T$ and $k_{8}^{\prime}$ as placeholders.