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Difference between revisions of "2009 AMC 10A Problems/Problem 19"
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== Solution ==
 
== Solution ==
  
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The circumference of circle A is <math>200\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 multiple of the circumference of B, therefore the quotient must be an integer.
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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.
  
 
Thus, <math>\frac{200\pi}{2\pi \cdot r} = \frac{100}{r}</math>.
 
Thus, <math>\frac{200\pi}{2\pi \cdot r} = \frac{100}{r}</math>.
  
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Therefore <math>r</math> 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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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>.
  
 
  *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>.
 
  *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>.

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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