6.3. Six pennies lie on the table, forming a closed chain (i.e., the first penny touches the second, the second touches the third, and so on, the sixth touches the first). A seventh penny, also lying on the table, rolls without slipping along the outer side of the chain, touching each of the six pennies in the chain in turn. How many revolutions will this penny make before returning to its original position?

## QUADRILATERAL AND FOUR CIRCLES
6.3. Answer. 4 turns.

From figure $a$, it can be seen that during the time the moving coin, depicted with a dashed line, rolls along the arc $\alpha$ of the stationary coin with center $O$, it rotates by an angle of $2 \alpha$: in this figure, $M^{\prime} A^{\prime}$ is the new position of the radius $M A$, the radii $M A$ and $M^{\prime} A^{\prime \prime}$ are parallel, $\angle A^{\prime \prime} M^{\prime} B = \angle A O B = \alpha$, and since the arcs $A^{\prime} B$ and $A B$ are equal, $\angle B M^{\prime} A^{\prime} = \alpha$, therefore, the entire angle $\angle A^{\prime \prime} M^{\prime} A^{\prime}$, by which the moving coin rotates, is $2 \alpha$.

![](https://cdn.mathpix.com/cropped/2024_05_21_20eb9f6ef2c568605a63g-50.jpg?height=356&width=391&top_left_y=539&top_left_x=285)
a)

![](https://cdn.mathpix.com/cropped/2024_05_21_20eb9f6ef2c568605a63g-50.jpg?height=276&width=274&top_left_y=583&top_left_x=704)
b)

Fig. 6.3.

Now let's find the sum of the arcs (in degrees) consisting of points of the stationary coins that the moving coin touched while rolling along the chain. Let $O_{1}, O_{2}, \ldots, O_{6}$ be the centers of the stationary coins (see figure b). The sum of the arcs lying inside the hexagon $O_{1} O_{2} \ldots O_{6}$ is equal to the sum of its interior angles, i.e., $180^{\circ} \cdot 6 - 360^{\circ} = 720^{\circ}$. Indeed, the sum of the interior angles of any $n$-sided polygon, not necessarily convex, is $180^{\circ} n - 360^{\circ}$ (prove this yourself or read the proof, for example, in the book by D. O. Shklyarsky et al., "Selected Problems and Theorems in Planimetry," "Nauka," 1967, problem 108). The sum of the arcs lying outside this hexagon is $360^{\circ} \cdot 6 - 720^{\circ} = 1440^{\circ}$. From this, we need to subtract the sum of the arcs lying in the indentations between two adjacent coins, where the moving coin will not reach; in each of the six indentations, the sum of the two such arcs is $120^{\circ}$ (on figure 6, these arcs are shown with a bold line). Thus, the total sum of the arcs along which the coin will roll is $1440^{\circ} - 120^{\circ} \cdot 6 = 720^{\circ}$, and the desired number of turns is obtained by multiplying this value by 2 and dividing by $360^{\circ}$.