Question number 26. Any three vertices of a cube form a triangle. What is the number of all such triangles whose vertices do not all lie on the same face of the cube? So here I have a cube and let's illustrate the two types of triangles. We can have the triangle that does not have all its vertices in the same face, for example in black like this, and in blue we can have a triangle here that does have all its vertices in the same face of the cube. And then we have to distinguish between them and count them. So in total there are eight times seven times six possibilities. And this comes from picking a vertex for one vertex of the triangle of which there are eight available. Then after we pick the first vertex of the cube the next vertex for the triangle can be chosen among the remaining seven vertices and so on. The last one we have only six choices for the last one, but some of these are counted more than once. And why is that? That is because of the following observation. What I can do is I can draw here, let's say this is the center of this face and then the opposite side is again the center of that face. I will draw here a line through those points and then we can imagine that the cube is being rotated about that line like this. And so what happens every time I make a rotation like that about the face, the positions of these triangles will change and I will obtain some other triangle that is in blue either on the opposite face so the blue triangle would move to the face here over there like that. That's where the blue one would go and then the black one would also change orientation but I'm not going to draw that in to mess up the picture. But what we observe is that the triangles that remain, that to begin with have vertices in one face like the one in blue, they also remain in some other face but they do not leave that face and the triangles of the other type like the black one will essentially become a different triangle. So there are six rotations possible through six rotations, one about each of the six faces. The triangles not contained in the face repeat exactly six times. Okay, so there are instead we have to we have to divide by six, eight times seven possibilities of which exactly six times four lie in the same face. And that is because through rotation these are not not affected and so we have to subtract those. So all together we have eight times seven times six divided by six minus six times four so that's eight times seven minus six times four which gives us 56 minus 24 or a total of 32. And these are the unique triangles that all lie, that all do not lie on the same face of the cube. So all the triangles like the black one over here but we are not over counting them through rotation and we have also subtracted all the possible triangles of which there are 24 that all lie in the same face. So the total here comes out to 32 and that is answer C.

unique triangles

cube vertices

rotational symmetry

combinatorial geometry

triangle configurations