## [ Rectangles and Squares. Properties and Characteristics  Method of Coordinates

On the plane, a square $A B C D$ is given. Find the minimum of the ratio $\frac{O A+O C}{O B+O D}$, where $O-$ is an arbitrary point on the plane.
It is known that for any rectangle $ABCD$ and any point $O$ in the plane of this rectangle, $OA^2 + OC^2 = OB^2 + OD^2$.

We will prove that $\frac{OA + OC}{OB + OD} \geqslant \frac{1}{\sqrt{2}}$, or $\sqrt{2}(OA + OC) \geqslant OB + OD$. Indeed,

\[
\begin{gathered}
\sqrt{2}(OA + OC) \geqslant OB + OD \Leftrightarrow (\sqrt{2}(OA + OC))^2 \geqslant (OB + OD)^2 \Leftrightarrow \\
\Leftrightarrow 2(OA^2 + 2OA \cdot OC + OC^2) \geqslant OB^2 + 2OB \cdot OD + OD^2 \Leftrightarrow \\
\Leftrightarrow OB^2 + 4OA \cdot OC + OC^2 \geqslant 2OB \cdot OD
\end{gathered}
\]

The last inequality is obvious, since $OB^2 + OC^2 \geqslant 2OB \cdot OD$.

If the point $O$ coincides with vertex $A$, we get that $\sqrt{2}AC = 2AB$, or

\[
\frac{OA + OC}{OB + OD} = \frac{AC}{2AB} = \frac{\sqrt{2}AB}{2AB} = \frac{1}{\sqrt{2}}
\]

Therefore, the value $\frac{1}{\sqrt{2}}$ is achieved. Consequently, the minimum value of the fraction $\frac{OA + OC}{OB + OD}$ is $\frac{1}{\sqrt{2}}$.

## Answer

$\frac{1}{\sqrt{2}} \cdot$