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Whether for hard candies, medicines, chlorine tablets, etc it seems that many applications would benefit from a 3D shape that maintains its surface area as it dissolves.

Anton Petrunin's answer to the 2017 MathOverflow question "Solids with constant surface area during 'erosion'" suggested a sphere with a small hole drilled through the center. This is a pretty good answer as it transitions from an almost-sphere to a toroid by the end of its life (which maintains a high surface area even as its volume approaches zero). For a shape that remains a single, unbroken solid throughout, this is probably as good as it gets?

But what if we relax that constraint and allow the shape to break into pieces as it dissolves? Is there some kind of fractal shape perhaps that might allow an even more consistent surface area as the shape dissolves?

EDIT: We're still assuming uniform dissolving, so any material within epsilon of the outside will dissolve within delta_t, and the total time for the shape to dissolve shouldn't approach zero.

EDIT2: Here's the sketch of an idea in 2D:

Self-splitting card

The idea here is that the card will repeatedly split in half as it dissolves, due to the specially shaped holes that will fill with solvent.

But I guess along with such "hole filling" and "splitting" there will always be discontinuities in the surface area over time (?), so it's impossible to hold the surface area truly "constant"... Is that true?

vaebnkehn
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  • Consider the figure obtained from a square of side $2$ by replacing the side $1$ squares in the upper left and lower right corners by side $1-t$ squares touching the center of the side $2$ square, $0\le t\le1$. (Think of all the mentioned squares being aligned with the co-ordinate axes). As $t$ goes from $0$ to $1$, the area diminishes (the candy dissolves) but the perimeter remains a constant $8$; when $t$ reaches $1$, you have two side $1$ squares, joined at a single point. Now let each of those two side $1$ squares dissolve in a similar way, to four side $1/2$ squares (continued) – Gerry Myerson Nov 28 '23 at 06:19
  • (continued) still of total perimeter $8$, then to eight side $1/4$ squares, and so on. Area goes to zero, perimeter stays constant. It's harder to visualize, but you can start with a side $2$ cube and dissolve away four side $1$ cubes from the corners, leaving four side $1$ cubes joined at a point, keeping the surface area at $24$, and iterating. – Gerry Myerson Nov 28 '23 at 06:25
  • I guess I should have specified in the OP, but in this problem, the shape must be dissolved uniformly: any matter within epsilon of the outside will dissolve away within delta_t. The problem is indeed easy if we allow for the object to be composed of different materials or allow the "acid" to be selectively applied. Or am I misunderstanding your suggestion? – vaebnkehn Nov 28 '23 at 06:40
  • Is there a known answer for two dimensions? Or is @GerryMyerson's comment intended to address that? – Brian Tung Nov 28 '23 at 07:04
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    You should specify everything that you have in mind, vae, so users don't waste our time, and yours, with answers that don't meet your secret conditions. – Gerry Myerson Nov 28 '23 at 07:59

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