Band Around a Cube

You have a cube with an edge length of s, and an elastic rubber band. The idea is to stretch the band around the cube so that half the area of the surface of the cube is below the band, and half the area of the of the cube above it; and every bit of the rubber band is at the same height, that is, it is all on one horizontal plane.

How long is the band? Is it constant or does it vary depending on the orientation of the cube?

If the cube is on a face, the situation is simple enough. Reveal Hidden Content
But what if the cube stands on an edge?
What if it stands on one of its points?
Is there a generic solution of some sort?
What is the shortest / longest possible length of the band?

The band is obviously not the same length regardless of the orientation of the cube.

If the cube is on one edge at a 45* angle, then the band crosses the hypotenuse twice and follows two of the cube edges [L = s * (2 + 2 sqrt(2)) ]

It quickly becomes complicated if the orientation of the cube is not a rotation along a single axis.

Very true. If the cube sits one of its six faces, the horizontal plane that goes through the center point of the cube touches the surface of the cube along a curve in the shape of a square, with perimeter length 4s. If one edge is down and one edge is up, as you showed, the curve is a rectangle with perimeter length of 2 x (1 + sqrt(2)) s, or about 4.8 s. If the cube stands on one point, the perimeter has a length of about 4.2 s.

I wonder if 4s and 2(1+sqrt(2))s are the lowest and highest possible lengths. On a hunch I'd say yes.. but since the cube can rotate around more than one axis, proving that is not trivial.

Originally posted by talzamir
Very true. If the cube sits one of its six faces, the horizontal plane that goes through the center point of the cube touches the surface of the cube along a curve in the shape of a square, with perimeter length 4s. If one edge is down and one edge is up, as you showed, the curve is a rectangle with perimeter length of 2 x (1 + sqrt(2)) s, or about 4.8 s. ...[text shortened]... say yes.. but since the cube can rotate around more than one axis, proving that is not trivial.

how did you get the solution for it standing on its point, you must have some good visualization power?

If the cube stands on one vertex and a horizontal plane slices at it, at the top it touches a point first, then forms an equilateral triangle until it is low enough to meet three vertexes of the cube. Going lower still the plane and the surface of the cube intersect along a hexagon. At the exact middle the hexagon has six sides of equal length, each the hypotenuse of a triangle with sides 0.5s, the length of the side is sqrt(2) / 2, and since there are six of them the total length is 6 x sqrt(2) / 2 = 3 x sqrt(2), or about 4.2.