I have two coins $A$ and $B$. The diameter of coins A is $d_A=18$ mm. The smaller coin $B$ rolls around $A$ without sliding. Both coins have marks on the edge. At the beginning of the movement, the coins touch the marks (see the figure).
To make the labels match again the first time, coin B should make two turns around $A$.
Question. What can be the diameter of coin $B$, $d_B$?
My attempt. The length of the circumference $C_A$ is related to the diameter $d_A$ by: $C_A=\pi d_A$, and the length of the circumference $C_B$ is $C_B=\pi d_B$. We know that $2 C_B = C_A$, therefore, $C_B =\frac{ C_A}{2}$, next $\pi d_B = \pi \frac{d_A}{2}$, $d_B = \frac{d_A}{2}=18/2=9$mm.
But correct answer is $12$ mm. Where is my error?
$\endgroup$2 Answers
$\begingroup$$12$mm is just one possible answer. What we need is that $B$ makes an odd number of revolutions while circling $A$ twice. So this would mean that twice the circumference of $A$ will be $k$ times the circumference of $B$, where $k$ is odd (to avoid matching on the first time around).
So the options are that $B$ has $\frac 23, \frac 25, \frac 27, \frac 29$ etc. of the circumference of $A$, and hence also the same proportion of the diameter.
For an integer mm diameter for $B$, we have the options of $\frac 23\times 18=12$mm and $\frac 29\times 18=4$mm.
$\endgroup$ $\begingroup$If $2C_B=C_A$, then $B$ would have to go around $A$ just one time to 'match up'!
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