I was wondering whether someone could come up with the proof of this exercise I set myself: We have four different types of triangles to play with (unlimited amounts of each): one triangle which has 3 sides of length 1 one triangle which has 2 sides of length 1 and one of length square root 2 one triangle which has 2 sides of length square root 2 and one of side length 1 one triangle which has 3 sides of length square root 2 When I assumble 4 triangles together, what is the maximum amout of irregular tetrahedron (3-D shape) I can form. And what is the proof to make sure I have found the maximum? My answer so far is: Using: 4 triangles the same shape ----- you can make 2 irregular tetrahedrons. 3 triangles the same shape----- you can make 2 irregular tetrahedrons. 2 the same + 2 same (but different to the 2 others) ---- you can make 1 irregular tetrahedron. 2 the same + 2 different --- you can make 3 irregular tetrahedrons (I'm not sure on this one) 0 different ---- obviously 0 irregular tetrahedron. Does this make any sense?