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Difference between revisions of "2009 AMC 10A Problems/Problem 19"

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== Problem ==
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Circle <math>''A''A</math> has radius <math>100</math>. Circle <math>''B''</math> has an integer radius <math>r<100</math> and remains internally tangent to circle <math>A</math> as it rolls once around the circumference of circle <math>''A''</math>. The two circles have the same points of tangency at the beginning and end of cirle <math>''B''</math>'s trip. How many possible values can <math>''r''</math> have?
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<math>
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\mathrm{(A)}\ 4\
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\qquad
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\mathrm{(B)}\ 8\
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\qquad
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\mathrm{(C)}\ 9\
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\qquad
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\mathrm{(D)}\ 50\
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\qquad
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\mathrm{(E)}\ 90\
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\qquad
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</math>
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== Solution ==
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The circumference of circle A is 200<math>\pi</math>, and the circumference of circle B with radius <math>r</math> is <math>2r\pi</math>. 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.
 
The circumference of circle A is 200<math>\pi</math>, and the circumference of circle B with radius <math>r</math> is <math>2r\pi</math>. 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.
  

Revision as of 11:41, 2 July 2009

Problem

Circle $''A''A$ 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.

$So\qquad\frac{200\pi}{2\pi*r} = \frac{100}{r}$

R must then be a factor of 100, excluding 100 (because then circle B would be the same size as circle A). $100\: =\: 2^2\; *\; 5^2$. Therefore 100 has $(2+1)\; *\; (2+1)\;$ 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 $a^x\: *\: b^y\: *\: c^z\;...$ and so on, is $(x+1)(y+1)(z+1)...$.
2009 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions
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