$23 \cdot 19$ In all the triangles formed by any 3 of the 8 vertices of a rectangular parallelepiped, the number of acute triangles is
(A) 0 .
(B) 6 .
(C) 8 .
(D) 24 .
(China High School Mathematics League, 1989)
[Solution] It is easy to know that, among the 8 vertices of a rectangular prism, any 3 vertices can form a triangle, totaling $C_{8}^{3}$ triangles, and these triangles can only be right-angled triangles or acute-angled triangles.
For each vertex of the rectangular prism as the right-angle vertex, there are 6 triangles, so among the $C_{8}^{3}$ triangles, there are $6 \cdot 8=48$ (triangles) that are right-angled triangles.
Thus, the number of acute-angled triangles is $C_{8}^{3}-48=8$.
Therefore, the answer is (C).
Note: A brief explanation for the conclusion that "triangles formed by the vertices of a rectangular prism can only be right-angled or acute-angled triangles":

For triangles with one side being an edge of the rectangular prism, they are clearly right-angled triangles; while triangles without any side being an edge of the rectangular prism are acute-angled triangles, as shown in the figure, we have
$$
A B^{2}=x^{2}+y^{2}, \quad B C^{2}=x^{2}+z^{2}, \quad A C^{2}=y^{2}+z^{2} .
$$

By the cosine rule, we have
$$
\begin{array}{l}
x^{2}+y^{2}=\left(x^{2}+z^{2}\right)+\left(y^{2}+z^{2}\right)-2 \sqrt{x^{2}+y^{2}} \cdot \sqrt{y^{2}+z^{2}} \cos C, \\
\therefore \quad \cos C=\frac{z^{2}}{\sqrt{x^{2}+z^{2}} \cdot \sqrt{y^{2}+z^{2}}}>0 .
\end{array}
$$

Since $0^{\circ}<\angle C<180^{\circ}$,
then $\angle C$ is an acute angle. Similarly, $\angle A, \angle B$ are also acute angles.
Therefore, the answer is $(C)$.