A convex body is bounded by 4 regular hexagonal and 4 regular triangular faces. At each vertex of the body, 2 hexagons and 1 triangle meet. What is the volume of the body if the length of its edges is unit length?
We will show that there is only one solid that satisfies the conditions of the problem, and we can obtain this, for example, by cutting off each vertex of a regular tetrahedron with a side length of 3 units with the plane determined by the points one-third of the way along the edges emanating from the respective vertex (Figure 1).

# 1988-10-311-1.eps 

Figure 1

First, we see that there can be at most one solid with the required properties. Consider, for instance, one of the hexagonal faces of the assumed solid and the 6 adjacent faces (those sharing an edge with it). These will alternate between triangular and hexagonal faces, since at each vertex, 2 hexagons and 1 triangle meet (Figure 2). If we construct a convex solid from the faces laid out in the manner shown, then after "folding" the faces, the points $A_{1}, A_{2}, A_{3}$, and $D_{1}$ and $D_{2}$ must coincide, as must the points $B_{1}, B_{2}, B_{3}$; $E_{1}, E_{2}$; $C_{1}, C_{2}, C_{3}$; and $F_{1}$, $F_{2}$.

## 1988-10-311-3.eps

Figure 2

The solid obtained by the previously described truncation of a regular tetrahedron with a side length of 3 units, due to the regularity of the tetrahedron, indeed has 4 regular triangular faces (one around each cut vertex) and 4 regular hexagonal faces (one on each face of the tetrahedron). It is also evident that in each of the 12 vertices, 2 hexagons and 1 triangle meet; thus, this solid indeed satisfies the conditions (Figure 3).

$$
\text { 1988-10-311-2.eps }
$$

Figure 3

After this, the volume of the solid can be easily calculated. The volume of a regular tetrahedron with side length $a$ is $\frac{a^{3} \sqrt{2}}{12}$; the volume of our solid is obtained by subtracting the volume of 4 regular tetrahedra with side length 1 unit from the volume of a tetrahedron with side length 3 units:

$$
V=\frac{27 \sqrt{2}}{12}-4 \frac{\sqrt{2}}{12}=\frac{23 \sqrt{2}}{12} \approx 2.71
$$