# Solution.

The height of the pyramid $SO = h$. The side of the base of the pyramid $AC = a$.

The height of the prism is $3h/4$, and the sides of the base of the prism are $a/2$.

The area of triangle $MNK$:

$$
S_{MNK} = \frac{a^2 \sqrt{3}}{16}
$$

The area of triangle $FPR$:

$$
S_{FPR} = \frac{a^2 \sqrt{3}}{64}
$$

The area of triangle $MPN$:

$$
S_{MPN} = \frac{1}{2} \cdot \frac{a}{2} \cdot \frac{3h}{4}, \quad S_{MPN} = S_{NPK} = S_{KRM}
$$

The area of triangle $FPM: S_{FPN} = \frac{1}{2} \cdot \frac{a}{4} \cdot \sqrt{\frac{9h^2}{16} + \frac{3a^2}{64}}, \quad S_{FPN} = S_{PRK} = S_{RFM}$.

Since the radius of the circle inscribed in triangle $ABC$ is $\frac{a \sqrt{3}}{6}$, and all lateral faces of the pyramid are inclined to the base plane at an angle $\alpha = \operatorname{arctg} \frac{3}{4}$, then $h = \frac{a \sqrt{3}}{8}$.

According to the problem, the total surface area of the polyhedron with vertices at points $M, N, K, F, P, R$ is $53 \sqrt{3}$, i.e.,

$$
\begin{gathered}
S_{\mathrm{MH}} = S_{MNK} + S_{FPR} + 3 S_{MPN} + 3 S_{FPN} = \\
= \frac{a^2 \sqrt{3}}{16} + \frac{a^2 \sqrt{3}}{64} + \frac{9ah}{16} + \frac{3a}{8} \sqrt{\frac{9h^2}{16} + \frac{3a^2}{64}} = \\
= \frac{a^2 \sqrt{3}}{16} + \frac{a^2 \sqrt{3}}{64} + \frac{9a^2 \sqrt{3}}{128} \\
\quad + \frac{3a}{8} \sqrt{\frac{27a^2}{16 \cdot 64} + \frac{3a^2}{64}} = \\
= \frac{a^2 \sqrt{3}}{16} + \frac{a^2 \sqrt{3}}{64} + \frac{9a^2 \sqrt{3}}{128} + \frac{15a^2 \sqrt{3}}{256} \\
= \frac{53a^2 \sqrt{3}}{16^2} = 53 \sqrt{3}
\end{gathered}
$$

Therefore, $a = 16$.

Answer: 16

![](https://cdn.mathpix.com/cropped/2024_05_06_fb302ee711d03f314b6cg-15.jpg?height=1210&width=874&top_left_y=163&top_left_x=1136)
9. The lateral face of a regular triangular pyramid $S A B C$ is inclined to the base plane $A B C$ at an angle $\alpha=\operatorname{arctg} \frac{3}{4}$. Points $M, N, K$ are the midpoints of the sides of the base $A B C$. Triangle $M N K$ is the lower base of a right prism. The edges of the upper base of the prism intersect the lateral edges of the pyramid $S A B C$ at points $F, P$, and $R$. The total surface area of the polyhedron with vertices at points $M, N, K, F, P, R$ is $53 \sqrt{3}$. Find the side of triangle $A B C$. (16 points)