Two coins puzzle

Place two identical coins side by side and roll one along the circumference of another without slipping. (Does it remind you of the way to draw epicycloids?). Two very related questions may now be asked:

The puzzle is actually an old-timer but never fails to surprise people who never saw it before. Even with the solution suggested by the picture on the right, it takes some experimenting before finally but still with the feeling of disbelief the right answer is accepted:

1. The coin will be the right side up.

2. It will take two revolutions for the rolling coin to get back to its starting position.

The puzzle illustrates how shaky our intuition about motion (and, by extension, intuition in general) may be. From observing drawing of the epicycloid it becomes apparent that the point on the rolling coin nearest to the other coin (i.e. the point where the two coins touch) will be the farthest from the latter after the rolling coin traverses a half-circumference. It's not yet a proof but an experiment that might help our intuition to set things straight.

The ultimate proof comes from the following observation. The rolling coin actually participates in two separate motions not unlike the moon relative to the earth:

• It rotates around its own center.
• It rotates around the center of the other coin.

To answer the first question, note that in both motions the moving coin makes a half-turn. A half-turn around a point can be described as a reflection in that point. Choose three points A,B,C on the rolling coin. When reflected in a point O, A is carried into A' while, B and C are carried into B' and C', respectively. Another reflection in a point O', moves A', B', C' into A", B", and C", respectively. Now note, that O is the middle point of the segments AA' and BB' whereas O' is the middle point of two segments A'A" and B'B". Therefore, in both triangles AA'A" and BB'B", the line OO' connects midpoints of two sides thus being parallel and equal to half size of the third side. From here, AA" and BB" are equal and parallel (both, of course, are equal and parallel to CC" as well.) Since points A,B,C have been chosen arbitrary on the moving coin, we can now say that the whole coin was translated along a vector AA".

Maxim R. made this remark:

The main point is that there is no slipping, so the point of contact will travel the distance Pi*r on both circles. If we label the rightmost point of the right coin A and rotate the right coin until it's on the left of the other coin, the point of contact will be A. That means the coin is the same side up as at the start.

References

1. J.R. Newman, The World of Mathematics, v. 3, Simon & Schuster Books, 2001