I think you want the *signed* total length of curvature-1 boundary, namely: (length of (convex) curvature-(+1) boundary) - (length of (concave) curvature-(-1) boundary) In which case Warren's proof will work, assuming that you are only allowed compact pieces with piecewise-C2 boundaries. I don't know what the definition of scissors-congruence is, but I imagine that it would be something like that (after all, can *you* cut out a shape that isn't compact or lacks a piecewise-twice-continuously-differentiable boundary using scissors???). Sincerely, Adam P. Goucher
----- Original Message ----- From: Allan Wechsler Sent: 04/29/14 11:06 PM To: math-fun Subject: Re: [math-fun] square-ball non-dissectability
The lemma is false, unless I have misunderstood it. Suppose you cut out a "biscuit" by making a cut along a circular arc of radius 1, centered on the circumference of the original circle. Now the amount of curvature-1 boundary has increased. Now glue the two pieces back together. You haven't made any progress, it's true, but at the second step the amount of curvature-1 boundary has decreased.
Something like this argument may go through, but this particular formulation looks _too_ facile to me.
On Tue, Apr 29, 2014 at 5:39 PM, Warren D Smith <warren.wds@gmail.com>wrote:
A square and circular disc of equal area are NOT inter-dissectable (i.e. scissors congruent, finite # pieces). Easy proof:
Lemma: The total length of curvature-1 boundary can never decrease as you cut up your disc and move the pieces around (gluing them at common boundary).
But it must decrease, in fact to 0, to achieve the conversion. Contradiction.
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