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November 23, 2003

More Cubic Bisection

pentagonal cubic bisection

The other day I asked:

  1. Can a cube be sectioned in such a way as to create a regular pentagon?

  2. It appears the regular hexagonal section has the greatest area of all possible sections.  Can you prove it?

I know you've been breathlessly waiting for the answers, so here you go.

First, no, a cube cannot be sectioned to create a regular pentagon.  The closest you can do is the figure shown above.  This is a "full house" pentagon; three of the sides are the same length, and the other two sides are the same length as each other, but longer than the other three.  {Note: it is not necessary that one of the pentagon's vertices be coincident with a vertex of the cube.}

Second, the regular hexagon is not the section with the greatest area.  I didn't mean for this to be a trick question, but I guess it was.  The section with the greatest area is this one:

maximal cubic bisection

The diagonal of each face is √2

Area: √2 = 1.41

Here's the regular hexagon again:

hexagonal cubic bisection

The diagonal of each face is √2

Each side of the hexagon is √2/2

Triangles are equilateral with area √3/8

Area: √3 = 1.30

There are some other candidates as well.  In the two figures above, consider rotating the section about the dashed line as an axis.  That yields the following section (a diamond, not a square):

diamondal cubic bisection

Each side is √5/2

One chord is √2, the other is √3

Area: √2√3 = 1.22

And continuing the rotation, this section, a square with the minimum area of any section which passes through the center of the cube:

square cubic bisection

Section same as cube face

Area: 1

Another interesting section is this one, the largest triangular section:

triangular cubic bisection

The diagonal of each face is √2

Triangles are equilateral with area √3/8

Area: √3 = 0.87

Finally, here's today's bonus question:

  • What is the area of the "full house" pentagonal section?

Posted by ole at 06:20 PM