I have seen various assertions that one can derive the isoperimetric inequality in the plane from Crofton's formula in geometric probability. Unfortunately, I have not managed to figure out such a proof (or find a reference containing said proof). Can anyone point me in the right direction?
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$\begingroup$ I don't think I've ever seen an argument using just Crofton's formula, but there's a proof at the end of these slides which uses some more general results from geometric probability: math.utah.edu/~treiberg/IntGeomSlides.pdf $\endgroup$ – Paul Siegel Jul 23 '16 at 22:15
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See the below beautiful note of Chris Croke's.
Christopher B. Croke, MR 2361884 A synthetic characterization of the hemisphere, Proc. Amer. Math. Soc. 136 (2008), no. 3, 1083--1086 (electronic).]
answered Jul 24 '16 at 3:33
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A proof of the isoperimetric inequality using Crofton's formula is contained in these notes by Treibergs: http://www.math.utah.edu/~treiberg/isoperim/isop.pdf
See also the proof in Integral Geometry and Geometric Probability by Santalo.
answered Feb 22 '17 at 13:12
