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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.
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== Problem ==
  
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<math>So\qquad\frac{200\pi}{2\pi*r} = \frac{100}{r}</math>
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Circle <math>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 circle <math>B</math>'s trip. How many possible values can <math>r</math> have?
  
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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*. Subtracting 1 from 9, so as to exclude 100, the answer is <math>\boxed{8}</math>.
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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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*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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[[Category: Introductory Geometry Problems]]
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== Solution ==
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The circumference of circle <math>A</math> is <math>200\pi</math>, and the circumference of circle <math>B</math> with radius <math>r</math> is <math>2r\pi</math>. Since circle <math>B</math> makes a complete revolution and ''ends up on the same point'', the circumference of <math>A</math> must be a multiple of the circumference of <math>B</math>, therefore the quotient must be an integer.
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Thus, <math>\frac{200\pi}{2\pi \cdot r} = \frac{100}{r}</math>.
 +
 
 +
Therefore <math>r</math> must then be a factor of <math>100</math>, excluding <math>100</math> because the problem says that <math>r<100</math>. <math>100\: =\: 2^2\; \cdot \; 5^2</math>. Therefore <math>100</math> has <math>(2+1)\; \cdot \; (2+1)\;</math> factors*. But you need to subtract <math>1</math> from <math>9</math>, in order to exclude <math>100</math>. Therefore the answer is <math>\boxed{8}</math>.
 +
 
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*The number of factors of <math>a^x\: \cdot \: b^y\: \cdot \: c^z\;...</math> and so on, where <math>a, b,</math> and <math>c</math> are prime numbers, is <math>(x+1)(y+1)(z+1)...</math>.
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== See Also ==
 +
 
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{{AMC10 box|year=2009|ab=A|num-b=18|num-a=20}}
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{{MAA Notice}}

Latest revision as of 08:57, 8 June 2021

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 circle $B$'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 multiple of the circumference of $B$, therefore the quotient must be an integer.

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

Therefore $r$ must then be a factor of $100$, excluding $100$ because the problem says that $r<100$. $100\: =\: 2^2\; \cdot \; 5^2$. 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, where $a, b,$ and $c$ are prime numbers, is $(x+1)(y+1)(z+1)...$.

See Also

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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