# 1953 AHSME Problems/Problem 46

## Problem 46

Instead of walking along two adjacent sides of a rectangular field, a boy took a shortcut along the diagonal of the field and saved a distance equal to $\frac{1}{2}$ the longer side. The ratio of the shorter side of the rectangle to the longer side was: $\textbf{(A)}\ \frac{1}{2} \qquad \textbf{(B)}\ \frac{2}{3} \qquad \textbf{(C)}\ \frac{1}{4} \qquad \textbf{(D)}\ \frac{3}{4}\qquad \textbf{(E)}\ \frac{2}{5}$

## Solution

Let $x be the sides of the rectangle. The length of the diagonal is $\sqrt{x^2+y^2}$, and the length of the two adjacent sides is $x+y$. Then the distance the boy saves is $x+y-\sqrt{x^2+y^2}$. Setting this equal to $\frac12y$, we have $$x+y-\sqrt{x^2+y^2}=\frac12y$$ $$x+\frac12y=\sqrt{x^2+y^2}$$ $$x^2+xy+\frac14y^2=x^2+y^2$$ $$xy=\frac34y^2$$ $$\frac xy=\frac34,$$ so the answer is $\boxed{\textbf{(D)}\ \frac{3}{4}}$.

## See Also

 1953 AHSC (Problems • Answer Key • Resources) Preceded byProblem 45 Followed byProblem 47 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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 All AHSME Problems and Solutions

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