Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution)
Line 20: Line 20:
 
== Solution ==
 
== Solution ==
  
−
The circumference of circle A is <math>200\pi</math>, and the circumference of circle B with radius <math>r</math> is <math>2(\pi)(r)</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 <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 perfect factor of the circumference of B, therefore the quotient must be an integer.
  
 
<math>So\qquad\frac{200\pi}{2\pi \cdot r} = \frac{100}{r}</math>
 
<math>So\qquad\frac{200\pi}{2\pi \cdot r} = \frac{100}{r}</math>

Revision as of 11:12, 31 December 2015

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)
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_19&oldid=74139"