Suppose we have a unit cube in ${\mathbb{R}^3}.$
We want to count the total number of distinct right triangles formed by connecting vertices of the unit cube. I can see that the total number of right triangles simply on a face of the cube will be $4$ and since there are $6$ faces we have $4 \times 6 = 24$ right triangles on the faces alone. Of course there are others across the cube's diagonals. I think the total will be $48$, due a symmetry across the diagonals, but am not entirely sure of my reasoning here, so I may be wrong.
If someone could elucidate or provide a better/more rigorous answer it would be appreciated.
Best Answer
A key insight is that the only instance in which three vertices are selected that do not form a right triangle, is when those vertices form an equilateral triangle--i.e., no two vertices share an edge.
Since the cube circumscribes two distinct regular tetrahedra (cf. stella octangula), there are a total of $8$ such triangles. Since there are a total of $$\binom{8}{3} = 56$$ ways to choose any three distinct vertices, and each choice corresponds to a unique triangle, the number of right triangles is $56 - 8 = 48$.