How to minimize exposed surface for half a pie? I bought a large pie (of radius $R$). I cut off a half and gave it to my friend. This exposed an area or $2Rh$ --  where $h$ is the pie's thickness -- to air.  I watched one Numberphile video too many, and now I want to rearrange my half of the pie so that the exposed area is minimal. I can get $\sqrt{2}Rh$ like this:

Main question: Can I do better?
Alternative question:  What if I had this in mind already when I was cutting off my half?
That is, can I divide a pie into two equivalent parts such that one (and thus also the other) of the remaining parts can be rearranged to have even smaller exposed area?
(Here "equivalent" means "such that they can be in turn subdivided  into finite number of subparts which can be matched into equal pairs", equal, in turn,  being "movable to each other, with the motion matching up the pieces inherited from the original boundary of the pie".)
My Osgood-Denjoy–Riesz pie-cutting knife is being resharpened, so piecewise-smooth cuts only, please.
 A: Great question!
I will answer the two-dimensional case, where your cake is simply a disk of unit radius and you wish to minimize the perimeter not originally part of the disk's border. This encompasses all dissections performed with vertical cuts and rearrangements of the pieces on a flat surface.  (I'm normalizing to radius $1$, since the problem is obviously invariant up to scaling.)
By the isoperimetric inequality, any shape of area $\frac{\pi}2$ must have a perimeter of at least $\pi\cdot\sqrt{2}$, which is less than the length-$\pi$ amount of side frosting you have available. So the best we can hope for is $(\sqrt{2}-1)\pi\approx 1.301$ unfrosted perimeter; your $\sqrt{2}$ solution is already pretty close. (Note that this argument doesn't rely on anything other than having $\pi$ perimeter available to us, so making weirder cuts won't improve on this bound.)
However, it turns out that we can get as close to this limit as we would like. The idea is to make (something very close to) a circle of radius $1/\sqrt{2}$, and ensure that we use the frosted border to cover as much of this as we can. I will handwave some of these arguments, because I think excessive rigor would just muddy the intuitive idea, but you could formalize this with a lot of epsilon-delta hand-wringing and gross trigonometry if you really wanted to.
Our strategy is as follows:

*

*Pick some very large $N$, and place $N$ points equally spaced on the semicircular frosted boundary. Cut off the frosted bits by cutting along these $N$ points, as shown below for $N=7$ (removed regions and frosting in red):



*

*Build a convex polygon with $N-1$ sides that match the length of the cuts in the previous step, and some finite number of other sides ($O(N)$ more certainly suffice), arranged so that the result looks very close to a circle of radius $\pi/\sqrt{2}$ (with correspondingly similar perimeter) and has the same area as the grey region above. (You have more than enough degrees of freedom to make this happen easily - let me know if I should elaborate more on this point.)


*Apply the Wallace–Bolyai–Gerwien theorem to turn the grey region into our new circular-ish polygon using finitely many pieces. (Note that because the theorem only uses translation and rotation of the pieces, we don't have to worry about turning a piece upside down and smearing frosting on the floor.)


*Glue the red bits back onto our $N-1$ free sides, covering up (in the limit) $\pi$ of our perimeter and leaving (in the limit) only $(\sqrt{2}-1)\pi$ uncovered.
So if we restrict ourselves to making vertical cuts in the cake, the minimum unfrosted area is $(\sqrt{2}-1)\pi Rh$.

In the 3-dimensional case, things will depend on the height of the cake. If $h$ is very small, you can do some silly things: for instance, cut little beveled squares out of the top frosted part of the cake, then put $6$ such frosted squares together in a cube and put the leftovers inside the cube; they will at least not see the "outside air", though perhaps this is considered cheating. (I suspect one can use similar constructions to still have arbitrarily little surface area anywhere, though.)
Using the isoperimetric inequality in 3 dimensions for surface area, we can work out that with a radius of $1$ and a height of $h$, we will need an excess area of at least $\pi(3h)^{2/3}-\pi(h+1)$, which is positive between $x\approx 0.81521$ and $x\approx 5.4115$ (this is assuming the bottom of the cake is frosted; if not, the range of such $x$ would be a bit larger). So perfection cannot be achieved at least for medium-tall cakes.
In general, trying to prove any more complex lower bounds on $3D$ solutions seems pretty complicated.
