A rectangular sheet of paper has a side ratio of $1: \sqrt{2}$. We want to mark an equally wide border on the sheet, which encloses half the area of the paper. How many times the width of the border should be the length of the shorter side of the paper?
Let the sides of the rectangle be $y$ and $\sqrt{2} y$, and the width of the frame be $x$. The area of the paper: $\sqrt{2} y^{2}$. The area of the small rectangle left in the center is half the area of the paper: $\frac{\sqrt{2}}{2} y^{2}$.

Express this area in terms of the sides of the rectangle: $(\sqrt{2} y-2 x) \cdot(y-2 x)$. Thus, the following equation can be written:

$$
\frac{\sqrt{2}}{2} y^{2}=(\sqrt{2} y-2 x) \cdot(y-2 x)
$$

From this, two values for $x$ are obtained: $x_{1}=\frac{1+\sqrt{2}+\sqrt{3}}{4} y$ (which is approximately 1.04y), and $x_{2}=\frac{1+\sqrt{2}-\sqrt{3}}{4} y$ (which is approximately 0.17y). Since the width of the frame cannot be more than $0.5 y$ (otherwise it would cover the entire sheet), the solution can only be $x_{2}$.

Therefore, the width of the frame is $\frac{1+\sqrt{2}-\sqrt{3}}{4} \approx 0.17$ times the shorter side.