2. The octahedron $A B C D E F$ has a square $A B C D$ as its base, while the line $E F$ is normal to the plane determined by the square $A B C D$ and passes through its center. It is known that the sphere that touches all faces of the octahedron and the sphere that touches all lateral edges (i.e., $E A, E B, E C, E D, F A, F B, F C$ and $F D$) have the same center, and that the sphere that touches the lateral edges has a surface area that is $50\%$ larger. Find the ratio $\frac{E F}{A B}$.
2. Let $G$ and $H$ be the midpoints of sides $AB$ and $CD$, and $O$ be the center of the square $ABCD$. Let $r_{1}$ and $r_{2}$ be the radii of two spheres. Then $r_{1}$ and $r_{2}$ are, respectively, the radii of the circles inscribed in quadrilaterals $FGHE$ and $FACE$. Denote $OG=a$, $OA=a' = a\sqrt{2}$, $OE=b$, and $OF=b'$. The common center $S$ of the two observed spheres lies on the segment $EF$, and $GS$ is the angle bisector of $\angle EGF$, while $AS$ is the angle bisector of $\angle EAF$ (since $S$ is the center of the inscribed circles in $FGHE$ and $FACE$). From this, it follows that $\frac{EG}{FG} = \frac{ES}{FS} = \frac{EA}{FA}$,

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By OP $20184 \mathrm{~A} 2$, this (by cross-multiplication) reduces to $\sqrt{a'^2 a^2 + a'^2 b'^2 + b^2 a^2 + b^2 b'^2} = \sqrt{a'^2 a^2 + a'^2 b^2 + b'^2 a^2 + b'^2 b^2}$, i.e., $a'^2 b'^2 + b^2 a^2 = a'^2 b^2 + b'^2 a^2$, which is equivalent to $\left(a'^2 - a^2\right)\left(b'^2 - b^2\right) = 0$. Since $a' \neq a$, from the previous equality it follows that $b' = b$, and from this we get $S \equiv O$. Now we have $r_{1} = \frac{ab}{\sqrt{a^2 + b^2}}$ and $r_{2} = \frac{a' b}{\sqrt{a'^2 + b^2}} = \frac{\sqrt{2} ab}{\sqrt{2a^2 + b^2}}$ (which we obtained by calculating the heights from $O$ in $\triangle EOG$ and $\triangle EOA$ with a right angle at vertex $O$). From the condition of the problem, we have $\frac{r_{2}}{r_{1}} = \sqrt{\frac{3}{2}}$, i.e., $\sqrt{\frac{3}{2}} = \frac{\sqrt{2} \sqrt{a^2 + b^2}}{\sqrt{2a^2 + b^2}} = \sqrt{\frac{2 + 2\left(\frac{b}{a}\right)^2}{2 + \left(\frac{b}{a}\right)^2}}$, and from this we get $6 + 3\left(\frac{b}{a}\right)^2 = 4 + 4\left(\frac{b}{a}\right)^2$, i.e., finally $\frac{EF}{AB} = \frac{b}{a} = \sqrt{6 - 4} = \sqrt{2}$.