Kvant Math Problem 29

Let the radius of each coin in the chain be $r$.

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Problem

$n$ identical coins are lying on a table, forming a closed chain (see the figure). How many rotations will a coin $M$ of the same size make while rolling once along the outer side of the entire chain, as shown in the figure (coin $M = 2~\text{коп.}$)? How does the answer change if the coin $M$ has a radius differing by a factor of $k$ from the radius of each coin in the chain?

Exploration

Let the radius of each coin in the chain be $r$. First examine small values of $n$ when the rolling coin has the same radius.

For $n=1$, a coin rolls around one fixed equal coin. This classical situation gives two complete turns, not one. One turn comes from the traveled arc length $2\pi(2r)$ divided by the circumference $2\pi r$, and one extra turn comes from the fact that the line joining the centers rotates through $2\pi$.

For $n=2$, the centers of the fixed coins form a segment of length $2r$. The center of the moving coin follows the boundary of the parallel curve at distance $2r$. This boundary consists of two straight segments and two semicircles. Its perimeter equals

$$2\cdot 2r+2\pi(2r)=4r+4\pi r.$$

Dividing by the circumference $2\pi r$ gives

$$2+\frac2\pi.$$

This is not an integer, hence arc length alone cannot be the answer. The orientation of the rolling coin must contribute additional rotation.

A convenient description is obtained from the center of the moving coin. If the moving coin rolls without slipping along a curve, the number of turns relative to a fixed direction equals

  1. the traveled length divided by its circumference;
  2. plus the total turning of the outward normal to the curve.

For a simple closed convex curve the normal turns through $2\pi$ during one circuit. Hence one extra revolution is always added.

The centers of the $n$ fixed coins form a regular $n$-gon with side length $2r$. The center of the rolling coin moves along the outer parallel curve at distance $2r$. The perimeter of this trajectory equals the perimeter of the polygon plus $2\pi(2r)$:

$$2nr+4\pi r.$$

Hence the contribution from arc length is

$$\frac{2nr+4\pi r}{2\pi r} = \frac n\pi+2.$$

Adding the extra revolution gives

$$\frac n\pi+3.$$

For a rolling coin of radius $R=kr$, the same reasoning suggests replacing the offset distance $2r$ by $r+R=r(1+k)$ and the circumference by $2\pi R$. This yields

$$\frac{2nr+2\pi(r+R)}{2\pi R}+1 = \frac{n}{\pi k}+\frac{1+k}{k}+1 = 2+\frac1k+\frac{n}{\pi k}.$$

The step most likely to conceal an error is the relation between total rotation and the sum of the arc-length contribution and the turning of the normal. That must be proved carefully.

Problem Understanding

The centers of $n$ identical coins of radius $r$ form a closed chain. Since neighboring coins touch, their centers are the vertices of a regular $n$-gon with side length $2r$. A coin $M$ rolls without slipping along the outside of the entire chain and returns to its starting position.

The task is to determine how many complete rotations the rolling coin makes. Then the same question must be answered when the radius of the rolling coin is $R=kr$.

This is a Type C problem, because a specific numerical quantity must be determined.

The core difficulty is that the rotation count is not obtained merely by dividing the traveled distance by the circumference of the rolling coin. The direction of the tangent to the trajectory changes during the motion, producing an additional contribution.

The answer should be

$$3+\frac n\pi$$

for equal coins, and

$$2+\frac1k+\frac{n}{\pi k}$$

for radius ratio $k$.

Proof Architecture

Let $R$ denote the radius of the rolling coin and $r$ the radius of each fixed coin.

Lemma 1. The center of the rolling coin moves along the outer parallel curve at distance $r+R$ from the polygon formed by the centers of the fixed coins; this follows from the constant distance between the centers of two tangent coins.

Lemma 2. The perimeter of that trajectory equals

$$2nr+2\pi(r+R);$$

the parallel curve consists of copies of the polygon sides shifted outward and circular arcs whose total angle is $2\pi$.

Lemma 3. If a circle of radius $R$ rolls without slipping along a smooth curve, then its total angle of rotation equals

$$\frac{L}{R}+\Theta,$$

where $L$ is the length traversed by its center and $\Theta$ is the total turning angle of the outward normal to the trajectory; this follows from decomposing the infinitesimal rotation into translation and change of tangent direction.

Lemma 4. For any simple closed convex curve, the outward normal rotates through $2\pi$ during one circuit; in particular, $\Theta=2\pi$ for the trajectory of the center.

The hardest point is Lemma 3, because it is where the additional revolution arises.

Solution

Let the radius of each coin in the chain be $r$, and let the radius of the rolling coin be $R$.

The centers of the fixed coins form a regular $n$-gon whose side length equals $2r$, since adjacent coins touch. Denote this polygon by $P$.

Since the rolling coin always touches the chain externally, the distance from its center to the point of contact is $R$, while the distance from the center of a fixed coin to the same point is $r$. Hence the center of the rolling coin remains at distance $r+R$ from the polygon $P$.

Thus the center moves along the outer parallel curve of $P$ at distance $r+R$.

The perimeter of $P$ is

$$2nr.$$

When a convex polygon is replaced by its outer parallel curve at distance $d$, each side contributes the same length as before, while each vertex is replaced by a circular arc of radius $d$. The sum of the exterior angles of a convex polygon is $2\pi$, so the total length of all these arcs is

$$2\pi d.$$

With $d=r+R$, the length of the trajectory of the center is therefore

$$L=2nr+2\pi(r+R).$$

Now consider the rolling motion. Let $ds$ be an infinitesimal displacement of the center along its trajectory. Because there is no slipping, the corresponding rotation caused by translation equals

$$\frac{ds}{R}.$$

If the tangent direction to the trajectory changes by an angle $d\varphi$, then the line joining the center of the rolling coin to the contact point rotates by the same angle. Relative to a fixed direction this contributes an additional rotation $d\varphi$ of the coin.

Hence the infinitesimal rotation angle of the coin is

$$d\alpha=\frac{ds}{R}+d\varphi.$$

Integrating over one complete circuit,

$$\alpha=\frac{L}{R}+\int d\varphi.$$

The trajectory of the center is a simple closed convex curve. During one complete traversal, its outward normal makes one full turn, so

$$\int d\varphi=2\pi.$$

Therefore

$$\alpha=\frac{L}{R}+2\pi.$$

The number of complete rotations is

$$N=\frac{\alpha}{2\pi} =\frac{L}{2\pi R}+1.$$

Substituting the value of $L$,

$$N = \frac{2nr+2\pi(r+R)}{2\pi R}+1.$$

After simplification,

$$N = \frac{nr}{\pi R}+\frac{r+R}{R}+1.$$

Writing $R=kr$ gives

$$N = \frac{n}{\pi k}+\frac{1+k}{k}+1 = 2+\frac1k+\frac{n}{\pi k}.$$

For equal coins, $k=1$, and therefore

$$N = 2+1+\frac n\pi = 3+\frac n\pi.$$

Thus the rolling coin makes

$$\boxed{,3+\frac n\pi,}$$

rotations when its radius equals the radius of the coins in the chain, and in the general case

$$\boxed{,2+\frac1k+\frac{n}{\pi k},}.$$

Verification of Key Steps

The first delicate point is the perimeter of the center trajectory. A careless computation may count a circular arc of radius $r+R$ around each vertex and then multiply by $n$. The correct argument uses the sum of the exterior angles. If the exterior angles are $\beta_1,\dots,\beta_n$, then

$$\beta_1+\cdots+\beta_n=2\pi,$$

hence the total arc length is

$$(r+R)(\beta_1+\cdots+\beta_n) = 2\pi(r+R),$$

independent of $n$.

The second delicate point is the extra revolution. For $n=1$ and $R=r$, the trajectory is a circle of radius $2r$. Its length is $4\pi r$, so arc length alone gives only

$$\frac{4\pi r}{2\pi r}=2$$

rotations. Direct observation of a coin rolling around an equal coin yields three rotations. The missing one is exactly the contribution of the $2\pi$ turning of the normal.

A third place where errors occur is the generalization to radius ratio $k$. The offset distance is not $2r$ but $r+R=r(1+k)$. Replacing it incorrectly by $2r$ would make the answer independent of $k$ in the curvature term, which cannot be correct. For example, when $R$ becomes very large, the number of rotations should decrease, and the formula

$$2+\frac1k+\frac{n}{\pi k}$$

has this property.

Alternative Approaches

One may regard the rolling coin as carrying a marked point on its rim. During an infinitesimal displacement $ds$, the no-slip condition produces a rotation $ds/R$. Simultaneously, the radius from the center to the contact point rotates together with the normal to the trajectory. Summing these two infinitesimal effects and integrating yields the same formula without explicitly discussing tangent vectors.

Another approach uses the support function of a convex curve. The trajectory of the center is the Minkowski sum of the polygon and a circle of radius $r+R$. The perimeter of a parallel curve increases by $2\pi(r+R)$, while the total curvature of any closed convex curve equals $2\pi$. Combining these two classical facts gives the result in a compact geometric form.