Let's choose the dimensions of a regular quadrilateral pyramid so that the ratio of the radii of the circumscribed and inscribed spheres is the smallest.
Let the base of the pyramid be the square $ABCD$, the vertex opposite to it be $M$, the midpoints of edges $AD$ and $BC$ be $F$ and $G$ respectively, and the midpoint of $FG$ be $K$. Then, the center of the circumscribed sphere coincides with the center $O$ of the circle circumscribed around the diagonal section $MAC$, and its radius is $r=OA$; similarly, the main circle cut by the plane $MFG$ from the inscribed sphere is the incircle of the triangle $MFG$. Denoting the center of this circle by $J$, the radius is $\varrho=JK$. Let $AB=1$, and our task is to determine the value of the height $KM=x$ for which the ratio $y=\frac{OA}{JK}$ is the smallest.

Based on known relationships and differentiation,

\[
\begin{gathered}
r\left(=\frac{abc}{4t}\right)=\frac{\sqrt{2} \cdot MA^2}{2AC \cdot KM}=\frac{2x^2+1}{4x} \\
\varrho\left(=\frac{2t}{2s}\right)=\frac{FG \cdot KM}{1+2FM}=\frac{x}{1+\sqrt{4x^2+1}} \\
y=\frac{r}{\varrho}=\frac{2x^2+1}{4x^2}\left(1+\sqrt{4x^2+1}\right)=\left(\frac{1}{2}+\frac{1}{4x^2}\right)\left(1+\sqrt{4x^2+1}\right) \\
y'=-\frac{1}{2x^3}\left(1+\sqrt{4x^2+1}\right)+\left(\frac{1}{2}+\frac{1}{4x^2}\right) \frac{4x}{\sqrt{4x^2+1}}= \\
=\frac{4x^4-2x^2-1-\sqrt{4x^2+1}}{2x^3 \sqrt{4x^2+1}}
\end{gathered}
\]

The numerator is 0 when

\[
4x^4-2x^2-1=\sqrt{4x^2+1}
\]

Squaring both sides and considering only the positive real root,

\[
4x^4(4x^4-4x^2-1)=0, \quad x_0=\sqrt{\frac{1+\sqrt{2}}{2}} \quad(=1.099)
\]

The ratio can only have an extremum at $x_0$, where

\[
\sqrt{4x_0^2+1}=1+\sqrt{2} \quad \text{and} \quad y_0=1+\sqrt{2} \quad(=2.414)
\]

Considering the value of our function at $x=1$ and $x=x_0$:

\[
\frac{3(1+\sqrt{5})}{4} \quad(=2.427), \quad \text{or} \quad \frac{5}{2} \quad(=2.5)
\]

both are greater than $y_0$, so the ratio $\frac{r}{\varrho}$ has its minimum at $x_0$. From $x_0$, all other dimensions of the pyramid can be calculated.

László Horváth (Csurgó, Csokonai Vitéz M. Gymnasium, 4th grade)

Note. At the height $x_0$, $r=\frac{1}{2} \sqrt{\sqrt{2}+1}$, $\varrho=\frac{1}{2} \sqrt{\sqrt{2}-1}$, from which $r^2-\frac{1}{2}=\varrho^2$, $OA^2-KA^2=JK^2$, so the centers of the circumscribed and inscribed spheres coincide.