1. Let $A B C D E F G H$ be a rectangular cuboid. How many acute-angled triangles are formed by joining any three vertices of the cuboid?
(1 mark)
Let $A B C D E F G H$ be a rectangular cuboid. If any three vertices are joined, how many acute-angled triangles can be formed?
(1 point)
1. 8
1. A total of $C_{3}^{8}=56$ triangles can be formed by joining any three vertices of the cuboid. Among these, if any two vertices of a triangle are adjacent vertices of the cuboid, the triangle is right-angled. Otherwise, it is acute-angled. To see this latter statement, note that if the dimensions of the cuboid are $p \times q \times r$, then we can find from the cosine law that the cosines of the angles of such a triangle will be equal to $\frac{p^{2}}{\left(p^{2}+q^{2}\right)\left(p^{2}+r^{2}\right)}, \frac{q^{2}}{\left(q^{2}+p^{2}\right)\left(q^{2}+r^{2}\right)}$ and $\frac{r^{2}}{\left(r^{2}+p^{2}\right)\left(r^{2}+q^{2}\right)}$, which are all positive.
For each fixed vertex (say $A$), we can form 6 triangles which are right-angled at $A$ (two on each of the three faces incident to $A$). Thus the answer is $56-6 \times 8=8$.