1953 AHSME 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}

Full credit goes to MAA for authoring these problems. These problems were taken on the AOPS website.