[ Rectangles and Squares. Properties and Characteristics ]

On a plane, there is a square $A B C D$. 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,

$O A^2 + O C^2 = O B^2 + O D^2$

We will prove that $\frac{O A + O C}{O B + O D} \geqslant \frac{1}{\sqrt{2}}$, or $\sqrt{2}(O A + O C) \geqslant O B + O D$. Indeed,

$$
\begin{gathered}
\sqrt{2}(O A + O C) \geqslant O B + O D \Leftrightarrow (\sqrt{2}(O A + O C))^2 \geqslant (O B + O D)^2 \Leftrightarrow \\
\Leftrightarrow 2(O A^2 + 2 O A \cdot O C + O C^2) \geqslant O B^2 + 2 O B \cdot O D + O D^2 \Leftrightarrow \\
\Leftrightarrow O B^2 + 4 O A \cdot O C + O C^2 \geqslant 2 O B \cdot O D
\end{gathered}
$$

The last inequality is obvious, since $O B^2 + O C^2 \geqslant 2 O B \cdot O D$.

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

$$
\frac{O A + O C}{O B + O D} = \frac{A C}{2 A B} = \frac{\sqrt{2} A B}{2 A B} = \frac{1}{\sqrt{2}}
$$

Therefore, the value $\frac{1}{\sqrt{2}}$ is achieved. Consequently, the minimum value of the fraction $\frac{O A + O C}{O B + O D}$ is $\frac{1}{\sqrt{2}}$.

## Answer

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