Re: [math-fun] Puzzle to find a function D^2 -> [0,1]
Rich wrote: << . . .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?
Actually, it seems easy to verify this in-betweenness for the circle directly. a) For a point outside the circle, its two tangent segments have lengths that total more than the circle's arclength between them. This amounts to showing that tan(theta) > theta for 0 < theta < pi/2. I believe Archimedes was capable of verifying this inequality even without calculus and modern terminology for trig functions. b) For a side of the inscribed polygon, that must be shorter than the circle's arc between the side's endpoints. a) and b) together prove the in-betweenness of the circle's length w.r.t. any in- & circumscribed polygons. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele
On Thu, Aug 14, 2008 at 12:40 PM, Dan Asimov <dasimov@earthlink.net> wrote:
a) For a point outside the circle, its two tangent segments have lengths that total more than the circle's arclength between them. This amounts to showing that
tan(theta) > theta
for 0 < theta < pi/2. I believe Archimedes was capable of verifying this inequality even without calculus and modern terminology for trig functions.
I think that's the hard part -- how do you show that? How did Archimedes do it? (If nobody knows, once I finish unpacking some boxes I should be able to find my copy of his complete works and look it up myself) I agree that it's trivial for the inscribed circle, and that Archimedes knowing this for the circle is far from proving Cauchy's general statement. --Joshua Zucker
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Dan Asimov -
Joshua Zucker