2007 AIME II Problems/Problem 11
Problem
Two long cylindrical tubes of the same length but different diameters lie parallel to each other on a flat surface. The larger tube has radius 
and rolls along the surface toward the smaller tube, which has radius 
. It rolls over the smaller tube and continues rolling along the flat surface until it comes to rest on the same point of its circumference as it started, having made one complete revolution. If the smaller tube never moves, and the rolling occurs with no slipping, the larger tube ends up a distance 
from where it starts. The distance can be expressed in the form 
where 
and 
are integers and is not divisible by the square of any prime. Find 
Solution
If it weren’t for the small tube, the larger tube would travel 
. Consider the distance from which the larger tube first contacts the smaller tube, until when it completely loses contact with the smaller tube.
Drawing the radii as shown in the diagram, notice that the hypotenuse of the right triangle in the diagram has a length of 
. The horizontal line divides the radius of the larger circle into 
on the top half, which indicates that the right triangle has leg of 48 and hypotenuse of 96, a 
.
Find the distance across the circumferences of the circles. The sixty degree central angle indicates to take 
of the circumference of the larger circle (twice), while the 
central angle in the smaller circle indicates to take 
of the circumference. This adds up to 
.
The actual horizontal distance it takes can be found by using the s. The missing leg is equal in length to 
. Thus, the total horizontal distance covered is 
.
Thus, we get 
, and our answer is 
.
See also
| 2007 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
