1. Six identical regular quadrilateral pyramids and a cube are made of wood. The face of the cube is identical to the bases of the pyramids. Determine the ratio of the surface area of the cube and the body that results from gluing the bases of the pyramids to the faces of the cube, if the ratio of the volumes of these bodies is $1: 2$.
1. The surface of the resulting body is formed by all 24 lateral faces of the given six pyramids. Let $v$ be the height of these pyramids and $a$ the length of their base edge (which is the same as the edge of the given cube). From the problem statement, it follows that the volume of one pyramid is one-sixth of the volume of the cube, i.e., $\frac{1}{3} a^{2} v=\frac{1}{6} a^{3}$, from which $v=\frac{1}{2} a$. The lateral face of the pyramid is an isosceles triangle, for which the height $w$ from the main vertex (Fig.1) satisfies the Pythagorean theorem $w^{2}=v^{2}+\left(\frac{1}{2} a\right)^{2}$, and after substituting $v=\frac{1}{2} a$, we get $w=\frac{\sqrt{2}}{2} a$. Therefore, the area of the lateral face of the pyramid is $\frac{1}{2} a w=\frac{\sqrt{2}}{4} a^{2}$. The resulting body has a surface area 24 times larger, i.e., $6 \sqrt{2} a^{2}$, while the surface area of the cube is $6 a^{2}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_28151cc061c394c64bdfg-2.jpg?height=216&width=401&top_left_y=342&top_left_x=1485)

Fig. 1

Answer: The ratio of the surface areas of the original cube and the resulting body is $1: \sqrt{2}$.

Note. Under the given assumptions, the resulting body by gluing will have twelve identical faces (a rhombic dodecahedron). The identical pyramids mentioned have a total volume equal to that of the given cube and are obtained by dividing the cube into six identical pyramids with a common main vertex at the center of the cube.

For a complete solution, 6 points are awarded, of which 3 points are for determining the relationship $a=2 v$.