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Difference between revisions of "2001 AMC 8 Problems/Problem 16"
 
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<math>\text{(A)}\ \dfrac{1}{3} \qquad \text{(B)}\ \dfrac{1}{2} \qquad \text{(C)}\ \dfrac{3}{4} \qquad \text{(D)}\ \dfrac{4}{5} \qquad \text{(E)}\ \dfrac{5}{6}</math>
 
<math>\text{(A)}\ \dfrac{1}{3} \qquad \text{(B)}\ \dfrac{1}{2} \qquad \text{(C)}\ \dfrac{3}{4} \qquad \text{(D)}\ \dfrac{4}{5} \qquad \text{(E)}\ \dfrac{5}{6}</math>
  
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==Solution==
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==Solution #1==
  
 
The smaller rectangles each have the same height as the original square, but have <math> \frac{1}{4} </math> 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 <math> \frac{1}{2} </math> 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 <math> 4 \times 1 </math> and the larger rectangle has dimensions <math> 4 \times 2 </math>. The ratio of their perimeters is <math> \frac{2(4+1)}{2(4+2)}=\frac{5}{6}, \boxed{\text{E}} </math>
 
The smaller rectangles each have the same height as the original square, but have <math> \frac{1}{4} </math> 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 <math> \frac{1}{2} </math> 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 <math> 4 \times 1 </math> and the larger rectangle has dimensions <math> 4 \times 2 </math>. The ratio of their perimeters is <math> \frac{2(4+1)}{2(4+2)}=\frac{5}{6}, \boxed{\text{E}} </math>
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 +
==Solution #2==
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A simpler way to solve the problem is by brute-forcing it. We know that the side length of the original paper is <math>4</math> inches by <math>4</math> inches. The smaller paper would have a side length of <math>4</math> inches by <math>4</math> · <math>\frac{1}{4} = 1</math> inches, having a perimeter of <math>2</math>(<math>4 + 1</math>) = <math>2</math>(<math>5</math>) = <math>10</math> inches. The larger rectangle would have a side length of <math>4</math> inches by <math>4</math> · <math>\frac{1}{2} = 2</math> inches, having a perimeter of <math>2</math>(<math>4 + 2</math>) = <math>2</math>(<math>6</math>) = <math>12</math> inches. Thus, the answer is <math>\frac{10}{12} = \frac{5}{6}</math>, i.e. <math>\boxed{\text{E}}</math>
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2001|num-b=15|num-a=17}}
 
{{AMC8 box|year=2001|num-b=15|num-a=17}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 07:13, 14 November 2024

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]

$\text{(A)}\ \dfrac{1}{3} \qquad \text{(B)}\ \dfrac{1}{2} \qquad \text{(C)}\ \dfrac{3}{4} \qquad \text{(D)}\ \dfrac{4}{5} \qquad \text{(E)}\ \dfrac{5}{6}$

Solution #1

The smaller rectangles each have the same height as the original square, but have $\frac{1}{4}$ 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 $\frac{1}{2}$ 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 $4 \times 1$ and the larger rectangle has dimensions $4 \times 2$. The ratio of their perimeters is $\frac{2(4+1)}{2(4+2)}=\frac{5}{6}, \boxed{\text{E}}$

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 $4$ inches by $4$ inches. The smaller paper would have a side length of $4$ inches by $4$ · $\frac{1}{4} = 1$ inches, having a perimeter of $2$($4 + 1$) = $2$($5$) = $10$ inches. The larger rectangle would have a side length of $4$ inches by $4$ · $\frac{1}{2} = 2$ inches, having a perimeter of $2$($4 + 2$) = $2$($6$) = $12$ inches. Thus, the answer is $\frac{10}{12} = \frac{5}{6}$, i.e. $\boxed{\text{E}}$

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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