A convex body has six square and eight regular hexagonal faces. We know that no two of the square faces share a vertex. Determine the volume of the body, given that it has an edge of unit length.
Solution. We can reach any edge of a convex polyhedron from any other edge by passing through a sequence of edges, where adjacent members are edges of the same face; therefore, if the sought solid exists, then all its edges must be of unit length.

If we sum the number of edges of the faces forming the polyhedron, we get twice the number of edges of the polyhedron, since two faces meet at every edge of the polyhedron. Therefore, the polyhedron has

$$
e=\frac{6 \cdot 4+8 \cdot 6}{2}=36
$$

edges. According to Euler's polyhedron theorem, the number of its vertices is $c=36+2-14=24$. The eight hexagonal faces together have 48 vertices, and no two squares share a vertex, and at every vertex of the polyhedron, at least three faces meet, so no three hexagonal faces can share a vertex. Therefore, at every vertex of the polyhedron, exactly one square face and two hexagonal faces meet. This also implies that each square face forms the same angles with its six adjacent hexagonal faces, and that the angle between any two adjacent hexagonal faces is the same (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_5a6aa9c1ccf6db9ab6aag-1.jpg?height=344&width=469&top_left_y=777&top_left_x=817)

Figure 1

Since we are dealing with a convex polyhedron, these conditions uniquely determine the solid up to congruence.

We get such a solid if we cut off from a regular octahedron with edge length 3 at each vertex of the octahedron the pyramid whose base is the square formed by the points one-third of the way along the four edges emanating from that vertex (Figure 2). The resulting solid has a square face at each vertex of the original octahedron, while each original triangular face of the octahedron becomes a regular hexagon (Figures 3-4).

![](https://cdn.mathpix.com/cropped/2024_05_02_5a6aa9c1ccf6db9ab6aag-1.jpg?height=372&width=353&top_left_y=1451&top_left_x=867)

Figure 2

![](https://cdn.mathpix.com/cropped/2024_05_02_5a6aa9c1ccf6db9ab6aag-1.jpg?height=301&width=336&top_left_y=1877&top_left_x=884)

Figure 3

![](https://cdn.mathpix.com/cropped/2024_05_02_5a6aa9c1ccf6db9ab6aag-1.jpg?height=328&width=331&top_left_y=2237&top_left_x=886)

Figure 4

Every edge of the cut-off pyramids is of unit length, so two such small pyramids can be assembled into a regular octahedron with unit edge length. The body diagonals connecting the opposite vertices of these small octahedra are diagonals of a unit square, so their length is $\sqrt{2}$ (Figure 5). Each such body diagonal is perpendicular to the plane defined by the remaining four vertices, so the volume of such a small octahedron is $\frac{\sqrt{2}}{3}$. The volume of the original octahedron is 27 times this, so the volume of the convex solid in question is

$$
V=27 \cdot \frac{\sqrt{2}}{3}-3 \cdot \frac{\sqrt{2}}{3}=8 \sqrt{2} .
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_5a6aa9c1ccf6db9ab6aag-2.jpg?height=461&width=422&top_left_y=463&top_left_x=838)

Figure 5

Remarks. 1. The fact that at every vertex of the sought polyhedron exactly one square face and two hexagonal faces meet can also be shown without using Euler's polyhedron theorem. It is known that if at any vertex of a convex polyhedron we sum the angles of the faces meeting at that vertex, the sum is less than $360^{\circ}$. Since every angle of a regular hexagon is $120^{\circ}$, it immediately follows that no three hexagonal faces can share a vertex.

2. The solid in question belongs to the family of so-called semiregular polyhedra. We call polyhedra semiregular if their vertex figures are congruent and their faces are regular polygons, but unlike regular polyhedra, they have faces of different numbers of sides (in this case, squares and hexagons).