I completely agree, this is a nice pure mathematical proof.

Better still, and to formalise the argument into neat algebra: Make the both coins hexagons. If the large coin has a radius say 3 times the size of the small one then you can easily see that you for the small hexagon to traverse the large one it has to be moved from one side touching the larger hexagon to the next touching the larger hexagon 6*3 times. In 6*(3-1) of these moves the small hexagon rotates through 360/6 degrees. In the other 6 (when it his a corner of the large hexagon) it rotates through 360/6+360/6. Adding all these up gives 6*(3-1)*(360/6)+6*(360/6+360/6) = (6+1)*360 i.e. 6+1 rotations.

Think of N sides and the large coin having x times the radius of the small. The case above is N=6, x=3. The argument works though for any N and any x. And it works for circles too: Just let N tend to infinity while holding the radius of the polygon fixed. It also, with small modification, works for non-integer x and so is completely general.

I completely agree, this is a nice pure mathematical proof.

Better still, and to formalise the argument into neat algebra: Make the both coins hexagons. If the large coin has a radius say 3 times the size of the small one then you can easily see that you for the small hexagon to traverse the large one it has to be moved from one side touching the larger hexagon to the next touching the larger hexagon 6*3 times. In 6*(3-1) of these moves the small hexagon rotates through 360/6 degrees. In the other 6 (when it his a corner of the large hexagon) it rotates through 360/6+360/6. Adding all these up gives 6*(3-1)*(360/6)+6*(360/6+360/6) = (6+1)*360 i.e. 6+1 rotations.

Think of N sides and the large coin having x times the radius of the small. The case above is N=6, x=3. The argument works though for any N and any x. And it works for circles too: Just let N tend to infinity while holding the radius of the polygon fixed. It also, with small modification, works for non-integer x and so is completely general.