Can a cube have a pentagon cross-section?

Question

lessthanjen has posted a question earlier:

http://answers.yahoo.com/my/profile;_ylt=ArJkC52lM0a7uiqDwrgwonQCxgt.?show=583930ef1acf4f8839393250bc8cd24caa

Is it possible to slice a cube with a single plane, resulting in a regular pentagon cross-section? An equilateral triangle, square, and regular hexagon cross section is possible, but what about a regular pentagon? How, or why not? 5-sided polygon cross-sections are possible, the question is whether or not a regular on is possible.

"..a regular one..."

JJ, I'm still trying to confirm this 2D pentagon outline of a 3D projection of the tesseract. However, it's not hard to show that a 5-sided cross-section of the cube is possible. I will make this problem easier by saying that a regular pentagon is not possible, but what's an elementary proof of this?

Answer 1

Since the faces of cube are three pairs of parallel planes,
the sides of the polygonal cross-section corresponding to
each pair of of planes are parallel.

All pentagonal cross-sections are parallelograms with
one trunceated vertex.

Since regular pentagon has no parallel sides, it cannot
be fomed by any planar section of a cube.

Answer 2

If a plane slices a polyhedron, it divides the vertices of the polyhedron into connected groups V and W on either side of the plane. The vertices on the resulting polygon are the points of intersection with the plane of the edges running between V and W.

THE REALLY SHORT PROOF:
If there are 3 vertices in V, slicing a cube of side length s, then the plane slice will give you a pentagon. Consider the vertex in V that has only one edge, e, to W. The polygon's angle at the vertex on e can be at most arccos(-1/5) < 108º. QED


A FEW DETAILS:
Assume that the three vertices in V are A-B-C, and that A, C, and B are also connected to D and E, to D and F, and to G, respectively. Now let the plane, P, strike AD, CD, AE, CF, and BG in A', C', E', F', and G', respectively.

Let
THE SHORT PROOF:
Note that in configuration 1, In configuration 2, 3s² = 2(5/4)s² - 2(5/4)s² cos(K)
1/2 = -5/2 cos(K)
K = arccos(-1/5) < 108º.
QED



THE LONG VERSION:
Recall that cos(108) = 1-φ = -1/φ where φ = (1+√5)/2 is the golden ratio.

Case 1: If E' and F' are the same distance from E and F, respectively then E'F'² = 2s². Furthermore, L² = s² + d² for some d>=0. Then E'F'² = 2s² = 2L² - 2L²cosK. That is,
s² = L²(1-cosK) = L²(1 - (1-φ)) = φL². But L>s so this cannot be. (In other words, E'G'F' cannot be obtuse).

Case 2: If E' and F' are different distances from E and F, respectively, then let the positive difference between these two distances be d. Then E'F'² = 2s² + d², and L = s²+(d/2)². Applying the law of cosines again to E'F'² = 2L² - 2L²cos108 =>
2s² + d² = 2(s²+(d/2)²) - 2(s²+(d/2)²)(1-φ)
d²/2 = 2(s²+(d/2)²)/φ
φd²/4 = s²+d²/4
d² = 4φs²
But this is also a contradiction since d is less than s.
QED

Answer 3

There is a moment when a 4-cube (8-cell) Tesseract,
viewed as a 3-D projection performing a double rotation
about two orthogonal planes will form a 5-sided polygon
(pentagon) in outline on a 2-D (viewing) plane.

OK. regular pentagons.
I am not aware of a case where slicing a regular hexahedron
will produce a regular pentagon. Regular forms entail division
of the number of known elements (faces, edges, vertices ie.
F=6,E=12,V=8) by 2 to produce a number of equal sides in
forming regular triangles, cubes and hexagons on a 2-D
plane. That number could never be 5.

Answer 4

it's definitely b because you do get a square. I also thing it's a.