How many triangles of different size and shape can be formed using the vertices of a cube?

Answer 1

Consider a "unit cube", with all edges equal to 1 inch in length. Eight vertices - A, B, C, D, clockwise around the top, E, F, G, H on the bottom, with A directly above E, B directly above F, etc. Triangle Type 1 is completely confined to one face of the cube. The second and third points are adjacent (connected by an edge of the cube) to the first, but are opposite each other, but still on the same face. Two of the sides are edges of the cube, and therefore have a length of 1 inch. The third side is a diagonal drawn across one face of the cube, and has a length of √2 inches. This is a right triangle, and is also an isosceles triangle (the two sides adjacent to the right angle have the same length). The area of this triangle is 1/2 square inch. A typical triangle of this type is ABC. Triangle Type 2 has two vertices that are adjacent to each other (on the same edge of the cube), but the third point is the opposite vertex of the cube from the first point, and is the opposite vertex on the same face as the second point. One side is an edge of the cube and has a length of 1. The second side is a diagonal drawn across one face of the cube, and has a length of √2. The third side is a diagonal drawn between opposite vertices of the cube, and has a length of √3. This is also a right triangle, but not an isosoceles triangle, and therefore different from the first type. The area of this triangle is √2/2. A typical triangle of this type is ABG. Triangle Type 3 has three vertices that are opposite each other along the same face (though on three different faces). I.e., Vertices 1 and 2 are opposite each other along one face, 2 and 3 are opposite each other along another face, and 1 and 3 are opposite each other along a third face. All three sides have a length of √2. This is an equilateral triangle. The area of this triangle is √3/2. A typical triangle of this type is ACF.