Re: [math-fun] Puzzle to find a function D^2 -> [0,1]
Rich wrote: << Quoting Dan Asimov <dasimov@earthlink.net>:
on a theorem of Cauchy:
Theorem: Given two convex curves A,B in the plane with B lying inside A, then length(B) < length(A).
Is it fair to give this one to Cauchy? Archimedes must have assumed something like it to estimate the circumference of a circle as being between the in- & circumscribed polygons. Was Cauchy just bringing the rigor up to modern standards, or is some non-obvious insight required?
I don't know the history of this theorem -- and hadn't realized Archimedes knew it. (How did he show this in the rigor of his day?) (Cauchy's proof goes something like this: First, show that if C is a convex curve in the plane, and you average the lengths of its projections onto lines of all directions, then that average is just half the length of the curve. (This is proved by first proving it for an interval, then approximating a convex curve by a convex polygon.) Now the above Theorem follows readily. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele
On Thursday 14 August 2008, Dan Asimov wrote:
(Cauchy's proof goes something like this: First, show that if C is a convex curve in the plane, and you average the lengths of its projections onto lines of all directions, then that average is just half the length of the curve. (This is proved by first proving it for an interval, then approximating a convex curve by a convex polygon.) Now the above Theorem follows readily.
Remark 1. I was idly thinking about the relative merits of this proof and a mindless inductive one, and briefly thought the inductive one had the advantage that you could do it in your head without having to work through the calculations for the average projected length of an interval. But actually you don't, because all that's really required for the proof to work is that the average length is *proportional to* the length of the interval, and that's trivial. Remark 2. This is much the same argument that gives a quick solution to Buffon's needle problem. (But expressed in terms of the average number of intersections with a set of equally spaced parallel lines. And the best way to get the constant of proportionality is to think about a circle rather than an interval.) -- g
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Dan Asimov -
Gareth McCaughan