For the discussion of math. Duh.

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Generic Goon
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Postby Generic Goon » Thu Oct 22, 2009 8:11 pm UTC

From wikipedia, "If the original cube has edge length 1, its dual octahedron has edge length sqrt{2}."

I'm having a hard time visualizing how this would be the case. For example, using this: http://www.math.uiowa.edu/~goodman/alge ... e-octa.GIF

If the sides of that cube were 1 unit, then the distance from one corner to another on the same face would be sqrt(2). But that would only be connecting the centers of two faces, not the vertices. And because they are slanted inwards, the faces must be closer together than the length of the edges, so it looks like the an edge on the octahedron would have to be larger than sqrt(2).

Also, looking at it again, it seems like it would have to be twice that, 2sqrt(2). The diagonal on the cube is the distance between two faces @ halfway up one of the pyramids that makes up the octahedron, and so shouldn't it be 1/2 of the total distance from one vertex to another?

What am I missing?

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Re: Octahedron-Cube

Postby gmalivuk » Thu Oct 22, 2009 9:37 pm UTC

Are you sure the dual octahedron is the one that's outside the cube? Wikipedia seems to suggest that it's the one inside: "One can construct the dual polyhedron by taking the vertices of the dual to be the centers of the faces of the original figure. The edges of the dual are formed by connecting the centers of adjacent faces in the original."

Of course, that still doesn't give sqrt(2). Where do you see that number given?

Edit: I'd say the cube Wikipedia article needs to be more specific. It is sqrt(2) if you take the dual octahedron to be the one whose edges intersect the edges of the cube at their midpoints. But if it's the one entirely inside, it'd be half that, and if it's the one entirely outside, it'd be 3/2 that, I believe.
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Re: Octahedron-Cube

Postby polytope » Fri Oct 23, 2009 3:19 am UTC

I'm pretty sure there is no standard size for the dual of a polyhedron. Duality relates polyhedrons, but not their measures.
If the octahedron is circumscribed about the cube, as in the picture the OP linked, then the edge length is [imath]\frac{3}{\sqrt{2}}[/imath] (If I calculated correctly).
If the octaherdron is inscribed inside the cube, as in this picture (http://en.wikipedia.org/wiki/File:Dual_ ... hedron.svg), then the edge length is [imath]\frac{1}{\sqrt{2}}[/imath].

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Re: Octahedron-Cube

Postby jaap » Fri Oct 23, 2009 6:44 am UTC

polytope wrote:I'm pretty sure there is no standard size for the dual of a polyhedron.

Well, often the size is chosen such that the edges of the two polyhedra intersect (or in this case bisect) each other. That way taking the dual of the dual results in the original polyhedron rather than a smaller or larger one.
With this convention, the octahedron does have edge length sqrt(2) times the edge length of the cube.

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Re: Octahedron-Cube

Postby PM 2Ring » Fri Oct 23, 2009 12:05 pm UTC

Jaap wrote:Well, often the size is chosen such that the edges of the two polyhedra intersect (or in this case bisect) each other.
If we do this with a cube & octahedron, we can see how they relate to the rhombic dodecahedron.

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