Difference between revisions of "1953 AHSME Problems/Problem 46"

Latest revision as of 01:11, 4 January 2019

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 $x<y$ 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 $\boxed{\textbf{(D)}\ \frac{3}{4}}$ .

1953 AHSC (Problems • Answer Key • Resources)
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 so the answer is <math>\boxed{\textbf{(D)}\ \frac{3}{4}}</math>.
-== See Also ==
+==See Also==
-{{AHSME 50p box|year=1953|num-b=46|after=Last Question}}
+{{AHSME 50p box|year=1953|num-b=45|num-a=47}}
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Difference between revisions of "1953 AHSME Problems/Problem 46"

Latest revision as of 01:11, 4 January 2019

Problem 46

Solution

See Also