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I am wondering whether there is a ‘simple’ proof of the Isoperimetric Theorem in the plane, i.e. that any simple closed curve in $\mathbb{R}^2$ with length $L$ and enclosed area $A$ fulfils $$4\pi A \leq L^2\ .$$

I should clarify what I mean by ‘simple’: I am looking for something that does not use very advanced mathematics (i.e. something like undergraduate-level mathematics, perhaps), but is also relatively short and understandable.

In particular, I was wondering whether it can be obtained easily from some other standard results, e.g. from Complex Analysis (Cauchy Integral Theorem, Riemann Mapping Theorem), but, if possible, avoiding things like Green's Theorem and Fourier series.

If you're wondering about these oddly specific requirements, I'm thinking about how to formalise this in a theorem prover.

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  • $\begingroup$ Equality is when the most conves case, the Circle occurs. $\endgroup$ – Narasimham Dec 21 '17 at 17:35
  • $\begingroup$ Yes; I'm not quite sure what you're getting at. $\endgroup$ – Manuel Eberl Dec 21 '17 at 18:58
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    $\begingroup$ Odd that you think Riemann Mapping is more elementary than Green's Theorem. That dumbfounds me. The two standard elementary proofs I know use the two techniques you mentioned. But look at Santaló's book on Integral Geometry and Geometric Probability, pp. 27-36, for a proof due to Blaschke using elementary integral geometry. (It's also in Blaschke's text on geometry, but I no longer have it to give a precise reference.) $\endgroup$ – Ted Shifrin Dec 21 '17 at 19:50
  • $\begingroup$ Well, we already have a formalisation of the Riemann Mapping theorem, but Green's theorem is still a bit incomplete at the moment. Thanks for the reference! $\endgroup$ – Manuel Eberl Dec 21 '17 at 21:51
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    $\begingroup$ @JackD'Aurizio Yes, I've looked into this problem a bit more yesterday and I also think Brunn–Minkowski would be a very nice route to take. I'll definitely look at your notes. $\endgroup$ – Manuel Eberl Dec 22 '17 at 10:57
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In Dido's problem Lagrange multiplier $\lambda$ that connects maximum Area $A$ for given boundary length $L$ is recognized as a curve of constant radius as the parameter $\lambda =r$ for the Circle. The proof is given in all books of Variational calculus.

Consequently when parameter $r$ is eliminated between area and circumference $$ A = \pi r^2,\, L= 2 \pi r ,\,$$ we obtain $$ 4 \pi A = L^2 $$ as a relationship for the maximum condition.. the inequality $$ 4 \pi A < L^2 $$ for all possible variations falling out from the full circle optimal situation.

Hope it serves as a simple enough proof.

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  • $\begingroup$ @Manuel Eberl Oh, you were just referring to this then? $\endgroup$ – Narasimham Dec 22 '17 at 20:29

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