Difference between revisions of "2009 AMC 10A Problems/Problem 19"
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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. | 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. | ||
| − | 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>. |
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>. | 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>. | ||
Revision as of 03:58, 31 January 2017
Problem
Circle 
has radius 
. Circle 
has an integer radius 
and remains internally tangent to circle as it rolls once around the circumference of circle . The two circles have the same points of tangency at the beginning and end of circle 's trip. How many possible values can 
have?
Solution
The circumference of circle A is 
, and the circumference of circle B with radius is 
. 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, 
.
Therefore must then be a factor of 100, excluding 100 (because then circle B would be the same size as circle A). 
. Therefore 100 has 
factors*. But you need to subtract 1 from 9, in order to exclude 100. Therefore the answer is 
.
*The number of factors ofand so on, where
and
are prime numbers, is
.
and so on, where 
and 
are prime numbers, is 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
