A rectangular parallelepiped has three edges of lengths $a, b, c$ meeting at one vertex. Establish a plane perpendicular to the body diagonal at each vertex and consider the convex body enclosed by these planes. What is the surface area and volume of this body?
We denote the vertices of the brick by $A, B, C, D, E, F, G, H$ (where $A B C D$ is one face and $A E\|B F\| C G \| D H$); let $a, b, c$ be the lengths of the edges $A B, A D, A E$ respectively. The length of the space diagonals is denoted by $d$, and the center of the brick by $O$; we know that $d^{2}=a^{2}+b^{2}+c^{2}$.

First, let's determine the shape of the solid $T$ formed by the planes perpendicular to the space diagonals. Reflecting the brick in the perpendicular bisector planes $T_{1}, T_{2}, T_{3}$ of the edges $A B, A D, A E$, the brick maps onto itself, the space diagonals map onto space diagonals, and thus the planes perpendicular to these diagonals also map onto each other, meaning that the solid $T$ is symmetric with respect to $T_{1}, T_{2}, T_{3}$.

Each of the octants created by the symmetry planes contains one vertex of the brick. The plane perpendicular to the corresponding space diagonal intersects the symmetry planes, forming a tetrahedron with one vertex at $O$, and the faces meeting at $O$ are pairwise perpendicular. The plane perpendicular to the space diagonal indeed intersects the rays that bound the octant, because the space diagonal passes through $O$, which is the common point of the rays, and forms an acute angle with the rays. Each such tetrahedron can be mapped to all others by reflecting in the symmetry planes, and the 8 tetrahedra together form $T$, whose boundary consists of 8 congruent triangles, with edges in the symmetry planes and vertices at their intersections.

![](https://cdn.mathpix.com/cropped/2024_05_02_9a25058d5d2834b1a953g-1.jpg?height=699&width=1174&top_left_y=716&top_left_x=451)

It is easy to show (which we will not use in our calculations), that the edges in one symmetry plane form a rhombus, and $T$ is composed of two pyramids based on such rhombi. Such a solid is called a rhombic bipyramid in crystallography.

The volume of $T$ is 8 times the volume of the tetrahedron between the symmetry planes, and its surface area is 8 times the area of the face of the brick passing through a vertex, denoted by $t$. Let the lengths of the edges of the tetrahedron parallel to the edges $A B, A D, A E$ of the brick be $j, k, l$ respectively. Then the volume of the tetrahedron can be calculated in two ways: as one-sixth of the product of the lengths of the perpendicular edges, and as one-third of the product of $t$ and half the space diagonal:

$$
\frac{1}{6} j k l=\frac{1}{6} d \cdot t
$$

Thus, for the volume $V$ and surface area $F$ of $T$,

$$
V=\frac{8}{6} j k l=\frac{4}{3} j k l \quad \text { and } \quad F=8 t=8 \frac{j k l}{d}
$$

Therefore, the following relationship holds between $F$ and $V$:

$$
F=\frac{6 V}{d}
$$

It is sufficient to determine one of $F$ and $V$. $V$ will be easier, and for its calculation, we need the segments $j, k, l$.

Let the vertex of the tetrahedron in the octant containing $A$ with the edge of length $j$ be $J$, and the intersection point of this edge with the face $A D H E$ (the center of this face) be $O_{1}$. Then $O O_{1}=a / 2$, and $J A$ is a line in the plane perpendicular to $O A$ at $A$, and thus perpendicular to $O A$. The right triangles $A J O$ and $O_{1} A O$ are similar because their acute angles at $O$ are equal; thus

$$
\frac{j}{d / 2}=\frac{O J}{O A}=\frac{O A}{O O_{1}}=\frac{d / 2}{a / 2}, \quad j=\frac{d^{2}}{2 a}=\frac{a^{2}+b^{2}+c^{2}}{2 a}
$$

Similarly, it can be shown that

$$
k=\frac{d^{2}}{2 b}, \quad l=\frac{d^{2}}{2 c}
$$

Thus, the desired quantities are

$$
\begin{aligned}
& V=\frac{4}{3} \cdot \frac{d^{6}}{8 a b c}=\frac{1}{6} \cdot \frac{\left(a^{2}+b^{2}+c^{2}\right)^{3}}{a b c} \\
& F=\frac{6 V}{d}=\frac{d^{5}}{a b c}=\frac{\left(\sqrt{a^{2}+b^{2}+c^{2}}\right)^{5}}{a b c}
\end{aligned}
$$