2009 AMC 10A Problem 19

As circle $B$ rolls without slipping, the arc of $A$ it has covered always equals the arc of $B$ that has been in contact. One trip around $A$ uses up arc length $2\pi \cdot 100$ on $A$, so the same total arc of $B$ has touched $A$.

The same point of $B$ is back at the contact point exactly when this arc is a whole number of $B$'s circumferences:
$$\frac{2\pi \cdot 100}{2\pi r} = \frac{100}{r} \text{ is an integer},$$
i.e. $r$ divides $100$.

$100 = 2^2 \cdot 5^2$ has $(2+1)(2+1) = 9$ positive divisors: $1, 2, 4, 5, 10, 20, 25, 50, 100$. The condition $r < 100$ removes $100$, leaving $8$ values.

The answer is $\boxed{\textbf{(B)}\ 8}$.

The geometry collapses to one sentence: rolling matches arc lengths, so "same point touching again" is a divisibility condition on circumferences, and circumference is proportional to radius. After that it is counting divisors, with a careful reading of the strict inequality $r < 100$.

• Counting all 9 divisors of 100 and forgetting the condition r < 100 excludes r = 100, or believing the answer must be 50 or 90.
• Worrying about the rotation of circle $B$ relative to the fixed frame (which differs by one turn for internal rolling); only the contact arc matters here.