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.
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.)
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 $D > 0$)?