How many distinct triangles can be constructed by choosing three vertices from among the corners of a unit cube?

Source: NCTM Mathematics Teacher, February 2008

**Solution**

We construct four triangles from three corners of a face for a total of triangles.

In the plane formed by the the two diagonals and and the two sides, we construct four triangles. Similarly, the diagonals C and D produce four more triangles. Thus, the Top/Bottom, Front/Back and Left/Right faces make total of triangles.

Take corner as the top and form a tetrahedron with the three diagonals as the base. The tetrahedron produces four triangular faces – three are repeat, one is distinct. Hence, eight corners make distinct triangles.

The number of distinct triangles constructed by choosing three vertices from among the corners of a unit cube = .

**Answer**:

**Alternative solution**

We construct a triangle from any three corners of the cube because they are not collinear. The number of ways to choose three corners from a set of eight = .