Question.Given a convex body $\Omega$, what is the shape of a surface $\Gamma$ of minimal area which divides $\Omega$ into two regions of equal volume?

### Background/motivation.

A 2D version of the question was posed by Michael Goldberg in *Monthly*: find the shortest curve which divides a convex quadrilateral into two equal areas. In the latter case the solution is given by a circular arc perpendicular to two sides of the quadrilateral (or just a segment of a straight line in degenerate situations). This follows from the observation that a circular sector is the shortest curve which, together with two sides of an angle, encloses a fixed area.

Goldberg himself offered a physically intuitive solution to the problem:

The curve may be considered as a restraining member under tension produced by internal fluid pressure in the restricted area. The ends of the curve are free to slide along the sides. Hence, the curve must be normal to two sides of the quadrilateral. Furthermore, since the fluid pressure is uniform, the curve must take the form of a circular arc.

~~The same approach suggests that in the general multidimensional case the solution is given by a spherical "cap" that intersects $\partial \Omega$ orthogonally. Now, assuming that the intuition is correct, is there a simple ~~*formal* proof of this conjecture?

**Edit.** As Sergei Ivanov points out the minimal surface in question need not be a spherical cap for dimensions $>2$.

**A modified question**: is there always at least one solution to the problem? Should one impose any smoothness conditions on $\partial\Omega$ to guarantee existence?