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

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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.
  
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<math>So\qquad\frac{200\pi}{2\pi*r} = \frac{100}{r}</math>
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<math>So\qquad\frac{200\pi}{2\pi \cdot r} = \frac{100}{r}</math>
  
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R must then be a factor of 100, excluding 100 (because then circle B would be the same size as circle A). <math>100\: =\: 2^2\; *\; 5^2</math>. Therefore 100 has <math>(2+1)\; *\; (2+1)\;</math> factors*. But you need to subtract 1 from 9, in order to exclude 100. Therefore the answer is <math>\boxed{8}</math>.
+
R must then be a factor of 100, excluding 100 (because then circle B would be the same size as circle A). <math>100\: =\: 2^2\; \cdot \; 5^2</math>. Therefore 100 has <math>(2+1)\; \cdot \; (2+1)\;</math> factors*. But you need to subtract 1 from 9, in order to exclude 100. Therefore the answer is <math>\boxed{8}</math>.
  
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*The number of factors of <math>a^x\: *\: b^y\: *\: c^z\;...</math> and so on, is <math>(x+1)(y+1)(z+1)...</math>.
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*The number of factors of <math>a^x\: \cdot \: b^y\: \cdot \: c^z\;...</math> and so on, is <math>(x+1)(y+1)(z+1)...</math>.
 
{{AMC10 box|year=2009|ab=A|num-b=18|num-a=20}}
 
{{AMC10 box|year=2009|ab=A|num-b=18|num-a=20}}

Revision as of 10:39, 27 October 2010

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.

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

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 $a^x\: \cdot \: b^y\: \cdot \: c^z\;...$ and so on, is $(x+1)(y+1)(z+1)...$.
2009 AMC 10A (Problems • Answer Key • Resources)
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Problem 20
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All AMC 10 Problems and Solutions
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