A rectangular parallelepiped has three edges meeting at a vertex: $a, b, c$, and $a>b>c$. On each face of the body, we place a roof-shaped body whose 4 sides are inclined at a $45^{\circ}$ angle to the base. How many faces, edges, and vertices does the resulting body have? What shapes are the faces? What is the volume of the body?
It is evident that each roof-shaped body (excluding the base, its edges, and vertices) has 4 side faces, 5 edges, and 2 vertices. One of the edges is parallel to the base, called the ridge - its endpoints are the vertices. The two longer base edges are connected by a symmetrical trapezoid, and the shorter ones by isosceles triangles; each side face is symmetrical to the perpendicular bisector plane of the corresponding base edge, and the angles on the base edge are equal.
![](https://cdn.mathpix.com/cropped/2024_05_02_b4d49f9f746cc43fc19cg-1.jpg?height=1070&width=618&top_left_y=358&top_left_x=735)

The foot of the perpendicular from the ridge vertices is at half the distance from the three edges of the base, compared to the shorter base edge, and the height of the roof is the same. Therefore, the length of the ridge is equal to the difference in the lengths of the base sides.

The side faces of the roofs placed on two adjacent faces of the rectangular prism, which rest on the common edge of the bases, lie on each other's extensions, because their dihedral angles are $45^{\circ}+90^{\circ}+45^{\circ}=180^{\circ}$. Thus, from the 24 faces of the 6 roofs, 12 faces remain on the body after assembly. Each face contains one edge of the rectangular prism diagonally, so every edge of the examined body belongs to only one roof, making the total number of edges $6 \cdot 5=30$. The vertices of the rectangular prism are also vertices of the examined body, so the total number of vertices is $8+6 \cdot 2=20$.

Symmetrical trapezoids connect to the longest ($a$-) edges of the rectangular prism through both of their faces, and when combined, they form hexagons with opposite sides parallel. The isosceles triangles connecting to the shortest ($c$-) edges similarly form rhombuses. Two triangular and trapezoidal faces connect to the $b$-edges, and when combined, they form axisymmetric pentagons. It is evident from the above that the 4 hexagons, rhombuses, and pentagons are congruent.

The roof placed on the $a$ and $b$ edges of the rectangular prism is divided by planes passing through the endpoints of the ridge and perpendicular to $a$ into a three-sided prism and two pyramids. When the latter are combined, they form a regular tetrahedron with a base edge of $b$ and a height of $b / 2$, so their combined volume is $b^{3} / 6$. The length of the lying prism is $a-b$, and its base (the cross-section perpendicular to its length) is an isosceles right triangle with a hypotenuse of $b$, so its volume is $b^{2}(a-b) / 4$. Therefore, the volume of the roof-shaped body is

$$
V_{a b}=\frac{b^{3}}{6}+\frac{b^{2}(a-b)}{4}=\frac{a b^{2}}{4}-\frac{b^{3}}{12}
$$

By substituting $a, b$ with $a, c$ and then $b, c$, we obtain the volumes $V_{a c}$ and $V_{b c}$ of the bodies placed on the intermediate and smallest faces of the rectangular prism, respectively. From these, the volume of the entire body is

$$
V=a b c+2\left(V_{a b}+V_{a c}+V_{b c}\right)=a b c+\frac{1}{2}\left(a b^{2}+a c^{2}+b c^{2}\right)-\frac{b^{3}}{6}-\frac{c^{3}}{3}
$$

Katalin Szabó (Csongrád, Batsányi J. g. IV. o. t.)