0
$\begingroup$

The puzzle is as follows:

Assume that you have three coins over a rectangular table as it is indicated in the picture from below. These three coins are identical in diameter which is 3 cm and tangent to each other. Suddenly a fourth coin coin equal to the previous ones in size and shape is made to roll tangentially around the others without sliding until reaching its original position. How many turns did this coin made?

Sketch of the problem

The given choices are:

  1. 3 turns
  2. 4 turns
  3. 1 turn
  4. 2 turns

This puzzle comes from a late 1970s APA IQ test timed cards adapted from Leon Thurstone and Catell's on psychometric measure.

I'm not sure if the question itself is okay or not. As the size of the table is given but I didn't use it.

What I tried to do was to calculate the length around those three coins and I concluded that it is $3\times 1.5 \pi$.

The number of turns given by that fourth coin would be:

$3\times 1.5 \pi\,cm\times\frac{1\,\textrm{turn}}{2\pi\times 1.5\,cm}$

This gives me:

1.5 turns. But it doesn't appear in any of the choices. What could be wrong here? Perhaps the size of the table contributes to the solution?

It would be helpful if answers included a drawing to spot which part I'm missing, because the more I look into this I can't find where I got it wrong.

$\endgroup$
4
$\begingroup$

The answer is

3 turns

The path around the three coins consists of three identical parts, so it is easiest to only look at one of those parts, i.e. a section during which it is in contact with one particular central coin.

The question says that the outer coin does not slip. To help our understanding, lets make the coin slip completely, so it always keeps the same point of its perimeter in contact with the central coin. The two extremal positions of this section of the path are on directly opposite parts of the central coin, so while the outer coin travels around from one side to the other it will perform a half turn. This is similar to how Australians are upside down relative to Europeans.
The distance travelled around the central coin is half its perimeter. If we don't let the outer coin slip during its trip, then the contact point on the outer coin must also have moved along half perimeter of that coin. This means that the coin must have made a further half turn. The coin therefore made a full turn on this section of its path, and hence three full turns in the full path.

The classic version with one central coin is called the Coin Rotation Paradox and it has been asked here before. This question is essentially half of that, times three.

There has also been a coin rolling question with other shapes.

$\endgroup$
2
  • $\begingroup$ Can you please add a drawing?. I attempted to visualize mentally what you mentioned but I still don't understand to which are you referring the rotation of the coin. Do you mean a slippage around a full coin or a rotation as it is indicated in the problem?. The size of the table is given but I think it is unneeded. All and all I still don't understand how come half turn becomes into one full rotation. $\endgroup$ – Chris Steinbeck Bell Mar 22 at 9:34
  • $\begingroup$ @ChrisSteinbeckBell See the linked wiki page for an animation with coins. $\endgroup$ – Jaap Scherphuis Mar 22 at 9:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.