Difference between revisions of "1989 AJHSME Problems/Problem 24"
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From the diagram, we can tell the [[perimeter]] of one of the small rectangles is <math>2(4x+x)=10x</math> and the perimeter of the large [[rectangle]] is <math>2(4x+2x)=12x</math>. The desired [[ratio]] is <cmath>\frac{10x}{12x}=5/6\rightarrow \boxed{\text{E}}</cmath> | From the diagram, we can tell the [[perimeter]] of one of the small rectangles is <math>2(4x+x)=10x</math> and the perimeter of the large [[rectangle]] is <math>2(4x+2x)=12x</math>. The desired [[ratio]] is <cmath>\frac{10x}{12x}=5/6\rightarrow \boxed{\text{E}}</cmath> | ||
| + | |||
| + | ==Solution 2 (Explaining Solution 1 Further)== | ||
| + | Let's assume that the side length is <math>6.</math> After we fold the square in half, the square will become two rectangles with side lengths <math>6</math> and <math>3.</math> After we cut the rectangle in half, the side lengths of the cut will be <math>6</math> and <math>1.5.</math> This will make the uncutted sides of the rectangle also <math>6</math> and <math>1.5.</math> But wait! There's one more of this on the other side. Because of this, we multiply the width by 2. So, the uncutted rectangle's perimeter will be <math>2(6+3) = 2(9) = 18.</math> Making it into a fraction, the ratio of one of the cut rectangles to the whole uncutted rectangle will be <math>\frac{15}{18} = \frac{5}{6}.</math> | ||
| + | |||
| + | ~DuoDuoling0 | ||
==See Also== | ==See Also== | ||
Revision as of 19:05, 12 December 2021
Problem
Suppose a square piece of paper is folded in half vertically. The folded paper is then cut in half along the dashed line. Three 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((0,0)--(0,8)--(6,8)--(6,0)--cycle); draw((0,8)--(5,9)--(5,8)); draw((3,-1.5)--(3,10.3),dashed); draw((0,5.5)..(-.75,4.75)..(0,4)); draw((0,4)--(1.5,4),EndArrow); [/asy]](http://latex.artofproblemsolving.com/6/a/b/6ab13c6ad222a415ca8ff4bfb6135d981e1c657c.png)
Solution
From here on, a blue line represents a cut, the dashed line represents the fold.
![[asy] draw((0,0)--(4,0)--(4,4)--(0,4)--cycle,linewidth(1)); draw((1,0)--(1,4),blue+linewidth(1)); draw((2,0)--(2,4),dashed); draw((3,0)--(3,4),blue+linewidth(1)); label("$x$",(0.5,4),N); label("$x$",(1.5,4),N); label("$x$",(2.5,4),N); label("$x$",(3.5,4),N); label("$4x$",(0,2),W); [/asy]](http://latex.artofproblemsolving.com/8/7/e/87eaf90020921c5f0b0a692a128d18a6aad5a735.png)
From the diagram, we can tell the perimeter of one of the small rectangles is 
and the perimeter of the large rectangle is 
. The desired ratio is ![\[\frac{10x}{12x}=5/6\rightarrow \boxed{\text{E}}\]](http://latex.artofproblemsolving.com/1/1/6/1166e492f670529bdb1c5b452f3f9ec2e43d95e5.png)
Solution 2 (Explaining Solution 1 Further)
Let's assume that the side length is 
After we fold the square in half, the square will become two rectangles with side lengths 
and 
After we cut the rectangle in half, the side lengths of the cut will be and 
This will make the uncutted sides of the rectangle also and But wait! There's one more of this on the other side. Because of this, we multiply the width by 2. So, the uncutted rectangle's perimeter will be 
Making it into a fraction, the ratio of one of the cut rectangles to the whole uncutted rectangle will be 
~DuoDuoling0
See Also
| 1989 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 23 |
Followed by Problem 25 | |
| 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 | ||
