TITLE: Solids with constant surface area during "erosion"
QUESTION [10 upvotes]: Imagine a drug, a pill that you swallow, which is designed to dissolve in your
stomach at a constant rate.  It must be shaped such that the surface area
remains constant when the volume is "eroded" uniformly over its surface.
Two dimensions
Define "erosion" of a distance $\delta$ from a shape as removing a slither of width $\delta$ around the whole perimeter of the shape.  This is a parallel curve, at least initially.  (Note that a re-entrant corner will become rounded after erosion.  An alternative defintion would have the corner preserved, which makes the question slightly easier but is harder to justify.)
Does a shape exist whose perimeter remains constant after is eroded (at least for erosion of a distance $\delta$, for all $\delta <= D$, for some $D > 0$)?
I have an unsatisfying solution:

 An annulus, assuming internal erosion is allowed, because the change
 of perimeter of the inner and outer circles cancel out until the area
 is zero.  This works
 mathematically but not for how I posed the question.  So I would be
 interested in solutions that don't have holes.

 But for an annulus with a channel of zero width connecting the inner
 and outer circle, after erosion the perimeter will have been reduced
 by about twice the width of the channel.  So I would conjecture that
 it can't be done without holes.

 (This solution doesn't generalise to spheres in 3D although holes
 might still help.)

Three dimensions
Define "erosion" of a distance $\delta$ from a solid as shaving a width $\delta$ from the whole surface of the solid.
Does a solid exist whose perimeter remains constant after is eroded (at least for erosion of a distance $\delta$, for all $\delta <= D$, for some $D > 0$)?

REPLY [10 votes]: It seems that such pill exists.
Take a ball and drill a hole through it, so you get a solid torus;
we assume it has smooth boundary $\Sigma$.
By Gauss--Bonnet formula, we gave
$$\int\limits_\Sigma G=0,$$
where $G$ denotes Gauss curvature.
Denote by $H$ the mean curvature of $\Sigma$;
it is mostly very negative in the surface of the hole.
It is easy to make a hole such that 
$$\int\limits_\Sigma H >0,$$
but, if the hole wriggles badly,  then
$$\int\limits_\Sigma H\approx -\infty.$$
It follows that for a reasonably wriggling hole, we get 
$$\int\limits_\Sigma G=\int\limits_\Sigma H=0.$$
Let $\Sigma_r$ be the $r$-equidistant surface from $\Sigma$.
Then by Weyl's formula
$$\textrm{area}\,\Sigma_r=\textrm{area}\,\Sigma+ r\cdot\int\limits_\Sigma H +r^2 \cdot\int\limits_\Sigma G$$
for sufficiently small $r$.
Therefore the identities above imply that the area of equidistant surfaces stays constant for a short time.