146 Among all the triangles formed by any three of the 8 vertices of a parallelepiped, the maximum possible number of acute triangles is
A. 28
B. 32
C. 34
D. 36
146 B. Let the parallelepiped be $A B C D-A_{1} B_{1} C_{1} D_{1}$. The number of triangles formed by three vertices is $C_{8}^{3}=$ 56. In the parallelogram $A B C D$, $\angle B A D$ and $\angle A B C$ are supplementary, so one of them must be $\geqslant 90^{\circ}$. Therefore, among $\triangle A B C$, $\triangle A C D$, $\triangle A B D$, and $\triangle C B D$, at least 2 are not acute triangles. The same applies to the parallelogram $A A_{1} C_{1} C$. Since there are $6 \times 2$ such parallelograms, at least $6 \times 2 \times 2=24$ triangles are not acute triangles. The number of acute triangles $\leqslant 56-$ $24=32$.
On the other hand, if each edge is 1, $\angle B A D=\angle B A A_{1}=\angle D A A_{1}=\alpha$, and $\alpha$ is an acute angle greater than $\arccos \frac{1}{3}$, then in this case, $\triangle B D A_{1}$ and 31 other triangles are all acute triangles.