67. Given a square $ABCD$ in the plane, find the minimum value of the ratio $\frac{OA+OC}{OB+OD}$. Here, $O$ is any point in the plane. (1993 St. Petersburg City Mathematical Selection Exam Problem)
67. First, prove $\frac{O A+O C}{O B} \geqslant \frac{1}{\sqrt{2}}$. Squaring both sides of the inequality and eliminating the denominator, we get
$$2\left(O A^{2}+O C^{2}+2 O A \cdot O C\right) \geqslant O B^{2} \mp O D^{2}+2 O B \cdot O D$$

Since $O A^{2}+O C^{2}=O B^{2}+O D^{2}$, the above inequality simplifies to
$$O B^{2}+O D^{2}+4 O A \cdot O C \geqslant 2 O B \cdot O D$$

This inequality clearly holds, because
$$O B^{2} + O D^{2} \geqslant 2 O B \cdot O D$$

Second, if we take $\theta=A$, we see that the value of the fraction is $\frac{1}{\sqrt{2}}$. Therefore, the minimum value of this ratio is $\frac{1}{\sqrt{2}}$.