1989 AJHSME Problems/Problem 24


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?

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

[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]


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]

From the diagram, we can tell the perimeter of one of the small rectangles is $2(4x+x)=10x$ and the perimeter of the large rectangle is $2(4x+2x)=12x$. The desired ratio is \[\frac{10x}{12x}=5/6\rightarrow \boxed{\text{E}}\]

Solution 2 (Explaining Solution 1 Further)

Let's assume that the side length is $6.$ After we fold the square in half, the square will become two rectangles with side lengths $6$ and $3.$ After we cut the rectangle in half, the side lengths of the cut will be $6$ and $1.5.$ This makes the perimeter of the cut rectangles $2(6+1.5) = 2(7.5) = 15.$ This will make the uncutted sides of the rectangle also $6$ and $1.5.$ 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 $2(6+3) = 2(9) = 18.$ Making it into a fraction, the ratio of one of the cut rectangles to the whole uncutted rectangle will be $\frac{15}{18} = \frac{5}{6}.$


See Also

1989 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
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All AJHSME/AMC 8 Problems and Solutions