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Obviously there are many applications for this and many solutions. I am also interested in closed curves that have the same ratio of area to arc length as we "grow" them, growth is done by augmentation in both cases.

THIS IS NOT ABOUT SCALING.

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    This is not possible: the ratio of surface area to volume is not "dimensionless". If you scale the object by λ, the ratio will grow accordingly.2011-05-09
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    @Fabian That just means that you cannot scale the solutions to this question. The question makes sense. For example, if you think of animals, their volume tells you how much heat they generate and their surface tells you how much heat they lose. So individual animals and species cannot grow by scaling. (Well, at least not for long without adapting a lot of other things.)2011-05-09
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    @user9325: I was reading "growth is done by augmentation" in terms of scaling. Of course, if you don't scale them then it can be done. But is the question then to characterize all curves who have the ratio 1 (in some units) of surface to volume?2011-05-09
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    I doubt there is anything that can be said about all such curves, I would just like a few elementary... Not too ugly examples... If such a thing exists.2011-05-09
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    @Fabian I think that the problem is to formulate the right question.2011-05-09
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    I tried, for example the Koch curve... It didn't work, but perhaps something like it can work.2011-05-09
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    @a little don: if you just scale things up it will *never* work, the intercept theorem (http://en.wikipedia.org/wiki/Intercept_theorem) tells you so.2011-05-09
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    I mention that in the original question, this isn't about scaling.2011-05-09

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An example: Start out with a square of size 1 and now grow one side as $x$ and one side as $y(x)$, then you want $(x+y)=2x y$ or $y(x)=\frac{x}{2x-1}$.

So after some time, one side stays constant $1/2$ and the object grows "only in one direction".

I doubt that you can get much more precise answers without having a more precise question, but it is possible that someone else knows the right question.

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    You did "beat me" by 12 seconds ;-)2011-05-09
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    This works, if we think of the augmented bit as an upside down "L" shape.2011-05-09
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Take a rectangle (with sides $a$ and $b$). The "volume" is given by $V=a b$; the "surface" is given by $S=2(a+b)$. Fix the ratio to $$r = \frac{S}{V} = \frac{2(a+b)}{ab}.$$ You can solve this equation for $$b = \frac{2a}{a r -2},$$ valid for $ar>2$. Plugging it into the expression for the volume, we obtain $$V= \frac{2 a^2}{a r -2}.$$ You can see that the volume grows arbitrarily large as $a\to2/r$.