Let $F_{1}, F_{2}, \ldots, F_{n}$ be one-pound coins affixed to the plane such that each $F_{i}$ touches $F_{i+1}$ $(i=1,2, \ldots, n$, and $F_{n+1}$ means $F_{1}$), and a similar one-pound coin $G$ can roll around the outside of this "chain" of coins, touching each $F_{i}$ in sequence. How many rotations does $G$ make about its own center as it returns to its starting position?
We denote the center of the bounding circle of $F_{i}$ by $O_{i}$. Due to the coins touching each other, each side of the closed polygon $O_{1} O_{2} \ldots O_{n}$ is equal to the diameter of the coins, which we choose as our unit of length. The constraint that the coins are fixed on the plane implicitly means that the coins do not overlap; for any pair of indices $i$ and $j$, $O_{i} O_{j} \geq 1$. Thus, the perimeter of the equilateral polygon $O_{1} \ldots O_{n}$ does not intersect itself, and we can speak of the direction of traversal of the polygon. Let the traversal of $O_{1}$, $O_{2}, \ldots, O_{n}$ be in the positive direction.

It is useful in describing the rolling to highlight those positions of $G$ where it is touching exactly two adjacent coins and is transitioning from resting on $F_{i}$ to resting on $F_{i+1}$. Let the center of the bounding circle of $G$ in this position be denoted by $G_{i+1}$, which we call the $(i+1)$-th resting point of the coin $G$. Then $O_{i} O_{i+1} G_{i+1}$ is an equilateral triangle with side length 1, so the resting points - and their bounding circles - can be pre-drawn.

![](https://cdn.mathpix.com/cropped/2024_05_02_2f870b711f16221b285eg-1.jpg?height=954&width=659&top_left_y=542&top_left_x=711)

1. Figure

According to the assumption, the triangle $O_{i} O_{i+1} G_{i+1}$ does not overlap the immediately preceding triangle $O_{i-1} O_{i} G_{i}$, and thus the sector containing $G_{i}$ and $G_{i+1}$ is at least $2 \cdot 60^{\circ}=120^{\circ}$ - otherwise, $G$ would jump directly from $F_{i-1}$ to $F_{i+1}$ without touching $F_{i}$ (2nd figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_2f870b711f16221b285eg-1.jpg?height=962&width=743&top_left_y=1731&top_left_x=672)

2. Figure

Thus, the internal angle of the $O$-polygon at $O_{i}$ is at most $240^{\circ}$.

The segments $O_{i} G_{i}$ and $O_{i} G_{i+1}$ cut out the points $K_{i}$ and $V_{i}$ from the circumference of $F_{i}$, between which $G$ will rest; $K_{i}$ will be the starting point, and $V_{i}$ the endpoint.

Let the reflection of the latter point in the tangent to $F_{i}$ at $K_{i}$ be $V_{i}^{\prime}$, then $V_{i}^{\prime}$ already marks the point on $G$ in the $i$-th resting point, with which it will rest on the endpoint $V_{i}$ of $F_{i}$ in the $(i+1)$-th resting point. The rolling, indeed, means that the arcs $K_{i} V_{i}$ and $K_{i} V_{i}^{\prime}$ are of equal length, and thus the central angles $K_{i} O_{i} V_{i}$ and $K_{i} G_{i} V_{i}^{\prime}$ are also equal. The former is positive as a rotation angle, the latter is negative. This means that the rotation angle $V_{i}^{\prime} G_{i} K_{i}=\beta_{i}$ is positive.

We claim that the rolling from the $G_{i}$ to the $G_{i+1}$ resting points will cause $G$ to rotate by twice the angle $K_{i} O_{i} V_{i}=\beta_{i}$. Indeed, the rolling consists of two rotations. If we rotate the circle around $G_{i}$ by $\beta_{i}$, it only touches $V_{i}$ at $K_{i}^{\prime}$. This is followed by the rotation around $G_{i+1}$ by the angle $V_{i}^{\prime} G_{i} K_{i}^{\prime}=\beta_{i}$, after which $V_{i}^{\prime}$ coincides with $V_{i}$.

It is clear that $\beta_{i}=360^{\circ}-2 \cdot 60^{\circ}-\alpha_{i}=240^{\circ}-\alpha_{i}$, where $\alpha_{i}=O_{i-1} O_{i} O_{i+1}$ is the internal angle.

When $G$ rolls around the chain of coins $F_{1} \ldots F_{n}$, each of its radii rotates by

$$
2\left(\beta_{1}+\beta_{2}+\ldots+\beta_{n}\right)=2 n \cdot 240^{\circ}-2\left(\alpha_{1}+\ldots+\alpha_{n}\right)
$$

angle. Since

$$
480^{\circ}=\frac{4}{3} 360^{\circ} \quad \text { and } \quad \alpha_{1}+\ldots+\alpha_{n}=(n-2) 180^{\circ}
$$

the rotation is the product of $360^{\circ}$ by

$$
\frac{4 n}{3}-(n-2)=\frac{n}{3}+2
$$

Thus, $G$ makes $2+n / 3$ revolutions around its center.

Our result is valid even for $n=2$, where both angles of the 2-sided polygon are $0^{\circ}$. In the case of $n=1$, as is well known, $G$ makes 2 revolutions as it rolls around $F_{1}$ (3rd figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_2f870b711f16221b285eg-2.jpg?height=786&width=1209&top_left_y=1320&top_left_x=447)

3. Figure

Remark. The fixing of the coins on the plane also means that each internal angle of the polygon $O_{1} O_{2} \ldots O_{n}$ is at least $60^{\circ}$, but this was not necessary for the calculation.