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Geometry Difficulty 2.5 Junior Find the answer

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 12\frac{1}{2} the longer side. The ratio of the shorter side of the rectangle to the longer side was:

Pick one

Solution

Let x<yx<y be the sides of the rectangle. The length of the diagonal is x2+y2\sqrt{x^2+y^2}, and the length of the two adjacent sides is x+yx+y. Then the distance the boy saves is x+y−x2+y2x+y-\sqrt{x^2+y^2}. Setting this equal to 12y\frac12y, we have
x+y−x2+y2=12yx+y-\sqrt{x^2+y^2}=\frac12y
x+12y=x2+y2x+\frac12y=\sqrt{x^2+y^2}
x2+xy+14y2=x2+y2x^2+xy+\frac14y^2=x^2+y^2
xy=34y2xy=\frac34y^2
xy=34,\frac xy=\frac34,
so the answer is (D) 34\boxed{\textbf{(D)}\ \frac{3}{4}}.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.