1953 AMC 12 Problem 46

Attempt Problem 46 of the 1953 AMC 12 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 1953 AMC 12 solutions, or check the answer key.

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

46.

Instead of walking along two adjacent sides of a rectangular field, a boy took a short-cut along the diagonal of the field and saved a distance equal to 12\frac{1}{2} the longer side. The ratio of the shorter side of the rectangle to the longer side was:

12\dfrac12

25\dfrac25

14\dfrac14

34\dfrac34

25\dfrac25

Answer: D
Concepts:rectanglePythagorean theoremratio
Difficulty rating: 1590
Small Hint:

Let the longer and shorter sides be LL and WW

Big Hint:

The condition is L+WL2+W2=L2L+W-\sqrt{L^2+W^2}=\frac{L}{2}

Solution:

The saving condition gives L2+W2=L2+W. \sqrt{L^2+W^2}=\frac L2+W. Squaring and canceling W2W^2 yields L2=L24+LW, L^2=\frac{L^2}{4}+LW, so WL=34.\frac{W}{L}=\frac{3}{4}.

Thus, the correct answer is D.

← Problem 45#45
Full Exam

Problem 46 in Other Years

1950 AMC 12 · 1951 AMC 12 · 1952 AMC 12 · 1954 AMC 12 · 1955 AMC 12 · 1956 AMC 12 · 1957 AMC 12 · 1958 AMC 12 · 1959 AMC 12