# 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?

• Even when bisecting the unit square, a straight line parallel to one pair of sides (length 1) is shorter than a quarter circle centered at a vertex (length 1.253...). – Rahul Aug 25 '11 at 5:01

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.

• I'm sorry but I don't understand the first sentence. What do you mean by "locally the surface must be a sphere" and how does that follow from the isoperimetric inequality? – t.b. Aug 25 '11 at 6:02
• This argument is beyond me. First you say that it must be a sphere "locally" and then you go ahead and consider all spheres cutting the cube in two equal parts intersecting the cube's surface orthogonally. That the plane then gives the minimal area among those is clear but I simply can't follow your reduction step. – t.b. Aug 25 '11 at 6:26
• @Theo: You're aiming at a slightly higher level of rigour than I was aiming at :-) Perhaps I should have said "we know from the proof of the isoperimetric inequality" or even just "we know because balloons are spherical". By "locally a sphere", I meant that any section of it must coincide with a section of a sphere; basically that it is a sphere, just not globally since it ends where the cube ends. (This is in response to your first comment; I just read your second comment now.) – joriki Aug 25 '11 at 6:32
• Okay, thanks for the explanation. Yes, I agree with what you outline in your comment (that's what I thought one should do), but I was of the impression that I missed something obvious. – t.b. Aug 25 '11 at 6:35
• @Theo: Sorry about that. I had understood the question to be mainly looking for "reasons" for reducing the complexity of searching through all the possible surfaces, and felt that optimal surfaces being spherical could be taken for granted without rigorous proof. Perhaps it would have been better not to mention the isoperimetric inequality at all; I just felt that the already plausible fact that a sphere has to be optimal locally would become even more plausible by reference to a theorem that it's optimal globally. – joriki Aug 25 '11 at 6:52

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.