Minimal area of a surface that splits a cube into two equal-size pieces I had read the following problem and its solution from one source problem which was the following:
You want to cut a unit cube into two pieces each with volume 1/2. What dividing surface, which might be curved, has the smallest surface area?
The author gave his first solution by this way:
When bisecting the equilateral triangle, an arc of a circle centered at a vertex had the shortest path. Similarly for this problem, the octant (one-eighth) of a sphere should be the bisecting surface with the lowest area. If the cube is a unit cube, then the octant has volume 1/2, so its radius is given by
$$\frac{1}{8}(\frac{4}{3} \pi r^3)=\frac{1}{2}$$
So the radius is $\displaystyle \left( \frac{3}{\pi} \right)^{\frac{1}{3}}$ and the surface area of the octant is
$$\text{surface area}=\frac{4 \pi r^2}{8}=1.52$$ (approximate)
But after this the author said that he made a mistake; the answer was wrong and the correct  one is the simplest surface – a horizontal plane through the center of the cube – which has surface area 1, which is less than the surface area of the octant. But he has not given reasons why the horizontal surface area is the best solution and I need a formula or proof of this. Can you help me?
 A: We know from the isoperimetric inequality that locally the surface must be a sphere (where we can include the plane as the limiting case of a sphere with infinite radius). Also, the surface must be orthogonal to the cube where they meet; if they're not, you can deform the surface locally to reduce its area. A sphere orthogonal to a cube face must have its centre on that face. You can easily show that it can't contain half the volume if it intersects only one or two faces. Thus, it must either intersect at least three adjacent faces, in which case its centre has to be at the vertex where they meet, or it has to intersect at least two opposite faces, in which case it has to be a plane.
A: Following up the discussion joriki and I had, rounding off the corners where the hexagon meets the edge of the cube does reduce the area.  If we look at the cross section where the hexagon meets the wall it looks like this:

$\theta$ is the dihedral angle between the hexagon and the cube.  I imagine rounding off the corner with a small radius R.  The surface area is reduced by $\frac{a}{\sqrt{2}}R[\cot \theta-(\frac{\pi}{2}-\theta)]$.  The area is increased by some small triangles, but their area is quadratic in $R$, being $R (\sec \theta -1)\sqrt{2}R(\frac{\pi}{2}-\theta)\frac{1}{2}$ so for small enough $R$ we have reduced the cut area.  We need to do the mirror image on the other side to maintain the volume, but we win there as well.
