Thread: 3-d visualization problem

A 3-d visualization problem. Thanks for the help

The centers of the faces of a cube are joined together to form an octahedron. The centrs of all faces of this octahedron are now joined together to form a smaller cube. What is the ratio of an edge of the smaller cube to an edge of the original cube?

Originally Posted by I-Think

A 3-d visualization problem. Thanks for the help

The centers of the faces of a cube are joined together to form an octahedron. The centrs of all faces of this octahedron are now joined together to form a smaller cube. What is the ratio of an edge of the smaller cube to an edge of the original cube?

Take the vertices of the cube to be the eight points (so its edges have length 2). Then the vertices of the octahedron will be the six points , , . Now work out the coordinates of the centres of the faces of the octahedron, and compare them with those of the original cube.

The point at which you will need to try to visualise the problem is that you need to decide which of the possible triples of vertices of the octahedron form the three vertices of one of the eight triangular faces. The key thing here is that opposite vertices of the octahedron cannot belong to the same face. So for example of one of the vertices of a face is (1,0,0), then that face cannot contain the vertex (–1,0,0).

Originally Posted by I-Think

A 3-d visualization problem. Thanks for the help

The centers of the faces of a cube are joined together to form an octahedron. The centrs of all faces of this octahedron are now joined together to form a smaller cube. What is the ratio of an edge of the smaller cube to an edge of the original cube?

If you look at one of the cube faces straight on you should get the view in the diagram below.

Now if you accept the fact that the height of the centre of a triangle is a third of the height of that triangle, then it should be easy to see that the side length of the inner cube will be a third of the side length of the outer cube.