2001 AMC 8 Problems/Problem 16
Contents
Problem
A square piece of paper, 4 inches on a side, is folded in half vertically. Both layers are then cut in half parallel to the fold. Three new rectangles are formed, a large one and two small ones. What is the ratio of the perimeter of one of the small rectangles to the perimeter of the large rectangle?
![[asy] draw((0,8)--(0,0)--(4,0)--(4,8)--(0,8)--(3.5,8.5)--(3.5,8)); draw((2,-1)--(2,9),dashed); [/asy]](http://latex.artofproblemsolving.com/7/e/9/7e91c72c703f0e511d42bca49f65aa22acae754b.png)
Solution #1
The smaller rectangles each have the same height as the original square, but have 
the length, since the paper is folded in half and then cut in half the same way. The larger rectangle has the same height as the original square but has 
the length, since the paper is cut in half after the fold but the fold retains both sides of the larger rectangle. Therefore, the smaller rectangles have dimensions 
and the larger rectangle has dimensions 
. The ratio of their perimeters is 
Solution #2
A simpler way to solve the problem is by brute-forcing it. We know that the side length of the original paper is 
inches by inches. The smaller paper would have a side length of inches by · 
inches, having a perimeter of 
(
) = (
) = 
inches. The larger rectangle would have a side length of inches by · 
inches, having a perimeter of (
) = (
) = 
inches. Thus, the answer is 
, i.e. 
See Also
| 2001 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 15 |
Followed by Problem 17 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
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