1989 AMC 8 Problem 24

Attempt Problem 24 of the 1989 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1989 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

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?
12\dfrac{1}{2}
23\dfrac{2}{3}
34\dfrac{3}{4}
45\dfrac{4}{5}
56\dfrac{5}{6}
Answer: E
Concepts:paper foldingperimeter
Small hint:
Give the original square a convenient side length, such as 4,4, and track the pieces
Big hint:
The piece containing the fold unfolds to a 2×42 \times 4 large rectangle, while each other layer is a 1×41 \times 4 small rectangle
Solution:
Let the square have side 4.4. Folding in half makes a 2×42 \times 4 stack of two layers. Cutting parallel to the fold splits it into a strip containing the fold and a strip that does not.
The strip with the fold unfolds into the large rectangle, 2×4,2 \times 4, with perimeter 2(2+4)=12.2(2 + 4) = 12. The other strip is two separate small rectangles, each 1×4,1 \times 4, with perimeter 2(1+4)=10.2(1 + 4) = 10. The ratio is 1012=56.\frac{10}{12} = \frac{5}{6}.
Thus, the correct answer is E .
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