n identical coins lie on the table, forming a closed chain. The centers of the coins form a convex polygon. How many revolutions will a coin of the same size make while rolling once along the outer side of the entire chain, as shown in the figure?

How will the answer change if the radius of this coin is $k$ times larger than the radius of each coin in the chain

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Let's assume the radius of the coins forming the chain is a unit. From Fig. 3, it can be seen that during the time a coin of radius $k$ rolls along the arc $\alpha$ of a stationary circle of radius 1, it rotates by an angle $\alpha(1+1 / k)$: in this figure, 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 $A^{\prime}$ В and $A B$ are equal in length, $\angle B M A^{\prime}=\alpha / k$, therefore, the entire angle $\angle A^{\prime \prime} M^{\prime} A^{\prime}$,

![](https://cdn.mathpix.com/cropped/2024_05_06_696cdbe703e904f1264dg-50.jpg?height=652&width=671&top_left_y=1549&top_left_x=1371)
by which the coin $M$ has rotated, is equal to $\alpha+\alpha / k$ (in particular, when $k=1$ this angle is $2 \alpha$). Now let's find the sum of the arcs consisting of such points of the stationary coins that the coin $M$ touched while rolling along the chain. Let $O_{1}, O_{2}, \ldots O_{n}$ be the centers of the coins in the chain. The sum of the arcs lying inside the polygon $O_{1} O_{2} \ldots O_{n}$ is equal to the sum of its internal angles, i.e., $\pi(n-2)$. The sum of the arcs lying outside the polygon, therefore, is $2 \pi n-\pi(n-2)=\pi(n+2)$. From this, we need to subtract the sum of the arcs lying in the depressions between two adjacent coins, which $M$ does not reach. In each of the $n$ depressions, the sum of two such arcs is $2 \pi / 3$ when $k=1$ (Fig. 4a) and $2 \arccos 1 /(k+1)$ in the general case (Fig. 4b). Thus, the sum of the arcs along which the coin $M$ will roll is $\pi(n+2)-2 \pi n / 3$ (in the general case, $\pi(n+2)-2 n \arccos 1 /(k+1))$. To find the required number of rotations, we need to multiply this value by 2 (in the general case, by $1+1 / k$) and divide by $2 \pi$. Answer. ($n / 3+2$) rotations when $k=1$; $[(k+1) / 2 k](n-2 / \pi n \arccos 1 /(k+1)+2)$ rotations in the general case ($n \geq 3$). In this solution, we used (when calculating the "losses" in the depressions between the coins) the fact that the coin $M$ rolls over all $n$ coins in a row, without exception. This condition is necessary for the problem to have a definite answer. For chains with "narrow gaps" (see fig.), of course, it is impossible to determine how much the coin $M$ will rotate based only on the number $n$. A sufficient, but not necessary, condition for the problem to be solvable is the following:
the polygon $O_{1} O_{2} \ldots O_{n}$ is convex. Think about how to obtain the answer if instead of a closed chain there is just one coin, two touching coins, and generally an open chain of $n$ coins $O_{1}, O_{2}, \ldots O_{n}$, where $O_{k}$ touches $O_{k+1}$, arranged so that $M$ can roll back and forth over all the coins in the following order: $O_{1}, O_{2}, \ldots O_{n-1}, O_{n}, O_{n-1}, \ldots O_{1}$ (see fig.) (Hint: It is sufficient to replace $n$ with $2 n-2$ in the formulas written above.)

## Answer

( [n/3]+2) rotations when $\mathrm{k}=1 ;[(k+1) / 2 k](n-[2 /(\pi)] n \arccos [1 /(k+1)]+2)$ rotations in the general case (n >= 3 ).