## Rolling coin Some perfectly circular coins are placed in a nearly circular chain. Each coin touches two other coins in the chain. There is room for least one more coin inside the chain.

A new coin is placed inside the chain and rolled around the chain once completely. The rolling coin touches at least one coin in the chain throughout the roll. It rolls without slipping. Then this coin is placed outside the chain and rolled around it in a similar manner in the opposite direction. All coins have the same radius. There is enough room for the rolling coin to touch each coin at least once in each roll.

How many times should we perform these two rolls so that the number of rotations made by the coin while rolling is twice the number of coins in the chain?

[SOLVED]

#### John Jones solved this puzzle:

I guess the "nearly" circular chain adds a bit of uncertainty to the puzzle. I assume only that the chain is convex and allows each rolling coin to touch each coin in the chain as it goes around.

To change the statement of the puzzle slightly but not significantly, I am going to have two rolling coins (rollers) and add their turns up for each move around a common coin. The rolling coins roll in opposite directions around the chain.

Look at a single move. A roll from one position to the next: the centre of the roller on the outside makes an equilateral triangle with the centres of the two starting coins. The roller on the inside ditto. And when they have finished the first move they are now tangent to the original (common) coin and the next in the chain - another pair of equilateral triangles. A diagram would help but is left as an exercise. ;-)

Look at the common coin. Imagine ink was applied to the edge of the rollers. The common coin will have ink over part of its edge on the outside and part on the inside. The angle subtended by the inked portion to the centre of the common coin is a measure of how much the rollers jointly turned.

If the coins were rolling on a flat line that would be the end of it, but the rollers are rolling around a circular coin and although I have never found this to be intuitive, the result is that the turn is doubled (if you roll a coin around another coin until it turns exactly once it will only be half way round).

The uninked portion of the common coin corresponds to the part of the coin inside the two pairs of equilateral triangles. Hence the angle subtended by the combined inked portion is 2π −3 since there are four 60° angles not inked.

So the joint number of turns is that divided by 2π and then doubled: (1 - 23) · 2 = 23.

So each move results in a joint turn of 23, giving 2n3 overall for each tour. The puzzle asks how many tours are needed to get a joint turn of 2n, which is clearly 2n / (2n3) = 3.

#### Susam Pal from cotpi added:

Let us see what happens when the rolling coin rolls over one coin inside the chain with the figure displayed below. The figure shows only a part of the chain with 3 adjacent coins. This is enough to determine how much the rolling coin rotates when it rolls over the coin in the middle. The rolling coin rolls over the coin centred at C1 and the centre of the rolling coin moves from R1 to R2. The arc of the circle centred at C1 enclosed within ∠R1C1R2 is the only part of the coin centred at C1 over which the rolling coin rolls. The rolling coin rotates by ∠R1C1R2 in the anticlockwise direction with respect to the line joining its centre and the point of contact with the arc on which it rolls.

However, it can be noticed that when the centre of the rolling coin moves from R1 to R2, the line joining its centre and the point of the contact with the arc of the coin in the chain rotates by ∠R1C1R2 in the anticlockwise direction with respect to the surface on which the coins are placed.

So, the rotating coin rotates by 2 · ∠R1C1R2 while rolling on the coin centred at C1 with respect to the surface on which the coins are placed. Now, from the figure, we get:

2 · ∠R1C1R2 = 2 · (∠CnC1C2 − ∠CnC1R1 − ∠C2C1R2)

The triangles in the figure are equilateral triangle because the length of each side of the triangles is twice the radius of a coin. So, the above equation can be rewritten as:

2 · ∠R1C1R2 = 2 · (∠CnC1C2 − 120°)

So, the rotating coin rotates by 2 · (∠CnC1C2 − 120°) while rolling on the coin centred at C1 with respect to the surface on which the coins are placed. After rolling over all the n coins in the chain, it rotates by:

2 · [(∠CnC1C2 − 120°) + (∠C1C2C3 − 120°) + … + (∠Cn−1CnC1 − 120°)]
= 2 · (∠CnC1C2 + ∠C1C2C3 + … + ∠Cn−1CnC1) − 2 · n · 120°
= 2 · (n − 2) · 180° − 2 · n · 120°

The last step follows from the fact that the sum of the interior angles of a simple n-gon is (n − 2) · 180°. Since, each 360° turn is equal to one rotation, the number of complete rotations made by the coin while rolling over the n coins in the chain is: (n − 2) − 2n3 = n3 − 2.

A similar analysis for the number of rotations made by the rotating coin while rolling outside the chain is n3 + 2.

We conclude that the total number of rotations made by the coin in both rolls is:

(n3 − 2) + (n3 + 2) = 2n3

So, we need to repeat these two rolls 3 times so that the coin rotates 2n times while rolling.

### Credit

This is an original puzzle from cotpi.