Problem 1427

1. Understanding the Problem:
We have $n \geq 3$ circular beer mats $B_1, B_2, \ldots, B_n$ of equal size arranged in a plane such that each mat $B_k$ touches $B_{k+1}$ and $B_{n+1} = B_1$. Another beer mat $B$ of the same size rolls along the outside of this chain of beer mats, touching each one in sequence. We need to determine how many rotations $B$ makes until it returns to its starting position.

2. Analyzing the Geometry:
Each beer mat $B_k$ touches $B_{k+1}$ at a single point. When beer mat $B$ rolls along the outside of the chain, it touches each beer mat $B_k$ along its circumference, except for the points of tangency.

3. Calculating the Total Arc Length:
The total circumference of one beer mat is $2\pi r$, where $r$ is the radius of the beer mat. Since $B$ rolls along the outside of $n$ beer mats, the total arc length covered by $B$ is $n \times 2\pi r$.

4. Considering the Points of Tangency:
At each point of tangency between two consecutive beer mats $B_k$ and $B_{k+1}$, $B$ does not cover a small arc. The angle subtended by this arc at the center of each beer mat is $60^\circ$ (since the tangency points form an equilateral triangle with the centers of the two touching beer mats).

5. Calculating the Uncovered Arc Length:
The angle subtended by the uncovered arc at each tangency point is $60^\circ$, which corresponds to an arc length of $\frac{60^\circ}{360^\circ} \times 2\pi r = \frac{\pi r}{3}$. Since there are $n$ tangency points, the total uncovered arc length is $n \times \frac{\pi r}{3} = \frac{n\pi r}{3}$.

6. **Total Arc Length Covered by $B$:**
The total arc length covered by $B$ is the total circumference of the $n$ beer mats minus the total uncovered arc length:
$$n \times 2\pi r - \frac{n\pi r}{3} = 2n\pi r - \frac{n\pi r}{3} = \frac{6n\pi r - n\pi r}{3} = \frac{5n\pi r}{3}$$

7. Number of Rotations:
The number of rotations $B$ makes is the total arc length covered divided by the circumference of $B$:
$$\frac{\frac{5n\pi r}{3}}{2\pi r} = \frac{5n}{6}$$
Since $B$ must make an integer number of rotations, we add 1 to account for the initial position:
$$1 + \frac{n}{6}$$