11.139. A cube is inscribed in a regular octahedron such that its vertices lie on the edges of the octahedron. By what factor is the surface area of the octahedron greater than the surface area of the inscribed cube?
Solution.

Square $ABCD$ is the common base of the regular pyramids $SABCD$ and $S_1ABCD$, which together form the given regular octahedron, $O$ is the center of this square, $MNPQM_1N_1P_1Q_1$ is the cube mentioned in the problem

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0849.jpg?height=729&width=640&top_left_y=43&top_left_x=108)

Fig. 11.36

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0849.jpg?height=456&width=462&top_left_y=321&top_left_x=770)

Fig. 11.37

$K$ is the center of the square $MNPQ$ - the intersection point of $SO$ and the plane of the square $MNPQ$ (Fig. 11.36). Let $AD=a, MQ=2b$. Then $MP=2b\sqrt{2}, AC=a\sqrt{2}, SO=AO=\frac{a\sqrt{2}}{2}, SK=SO-KO=\frac{a\sqrt{2}}{2}-b$. $\triangle SMP \sim \triangle SAC$. From the similarity, it follows: $\frac{MP}{AC}=\frac{SK}{SO}; \quad \frac{2b\sqrt{2}}{a\sqrt{2}}=\frac{\frac{a\sqrt{2}}{2}-b}{\frac{a\sqrt{2}}{2}}$; $2b\sqrt{2}=a\sqrt{2}-2b; 2b=\frac{a\sqrt{2}}{\sqrt{2+1}}=a\sqrt{2}(\sqrt{2-1})$. The surface area of the octahedron $S_1=8S_{\triangle ASD}=8 \cdot \frac{a^2\sqrt{3}}{4}=2a^2\sqrt{3}$. The surface area of the cube $S_2=6S_{MNPQ}=6 \cdot (2b)^2=$ $=6(a\sqrt{2}(\sqrt{2}-1))^2=12a^2(\sqrt{2}-1)^2=12a^2(3-2\sqrt{2})$

Then $\frac{S_1}{S_2}=\frac{2a^2\sqrt{3}}{12a^2(3-2\sqrt{2})}=\frac{\sqrt{3}(3+2\sqrt{2})}{6}$.

Answer: $\frac{\sqrt{3}(3+2\sqrt{2})}{6}$.

11.140. Find the volume of a regular triangular pyramid (Fig. 11.37), where the plane angle at the vertex is $90^\circ$, and the distance between a lateral edge and the opposite side is $d$.

## Solution.

We will assume that $B$ is the vertex of the pyramid, and $\triangle ASC$ is the base. $SD$ is the height of the right isosceles $\triangle ASC$, dropped to the hypotenuse $AC$. Then $SD$ is the common perpendicular to the skew lines $BS$ and $AC$, and its length is the distance between them, $SD=d$. Therefore, $AC=2SD=2d, S_{\triangle ASC}=\frac{1}{2}AC \cdot SD=d^2$, $BS=AS=SD\sqrt{2}$. The volume of the pyramid $V=\frac{1}{3}S_{\triangle ASC} \cdot BS=\frac{d^3}{3}\sqrt{2}$.

Answer: $\frac{d^3}{3}\sqrt{2}$.