2009 AMC 10A Problems/Problem 19

Problem

Circle $A$ has radius $100$ . Circle $B$ has an integer radius $r<100$ and remains internally tangent to circle $A$ as it rolls once around the circumference of circle $A$ . The two circles have the same points of tangency at the beginning and end of cirle $B$ 's trip. How many possible values can $r$ have?

$\mathrm{(A)}\ 4\ \qquad \mathrm{(B)}\ 8\ \qquad \mathrm{(C)}\ 9\ \qquad \mathrm{(D)}\ 50\ \qquad \mathrm{(E)}\ 90\ \qquad$

Solution

The circumference of circle A is $200\pi$ , and the circumference of circle B with radius $r$ is $2r\pi$ . Since circle B makes a complete revolution and ends up on the same point, the circumference of A must be a perfect factor of the circumference of B, therefore the quotient must be an integer.

Thus, $\frac{200\pi}{2\pi \cdot r} = \frac{100}{r}$

R must then be a factor of 100, excluding 100 (because then circle B would be the same size as circle A). $100\: =\: 2^2\; \cdot \; 5^2$ . Therefore 100 has $(2+1)\; \cdot \; (2+1)\;$ factors*. But you need to subtract 1 from 9, in order to exclude 100. Therefore the answer is $\boxed{8}$ .

*The number of factors of  and so on, where  and  are prime numbers, is .

2009 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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All AMC 10 Problems and Solutions

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2009 AMC 10A Problems/Problem 19

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