This statement puzzles me, since what I meant by being able to slide a polygon around a curve is that the vertices remain on the curve and the edges are rigid segments. (So during such a sliding each edge, and so the total perimeter, remains constant.) PUZZLE: Specify a case where a finite number of unit balls are each tangent to two others in a unit-radius tunnel* through R^3 (and not otherwise intersect) . . . but cannot be slid (i.e., so that each ball moves to the position of an adjacent one, through a continuous family of such cyclically tangent arrangements) -- and prove this. --Dan __________________________________________ * Meaning: there is a C^1 curve C in R^3 such that the tunnel is the union of all the unit-radius disks -- required to be disjoint -- that are each centered at a point of C and perpendicular to C. << . . . Given an ellipse, you can start my recursive perpendicular bisector subdivision from _any_ line through the center of the ellipse. So, it _is_ possible to "slide" the vertices around the perimeter of the ellipse, at least for equilateral 2^n-gons. Note, however, that the total perimeter of the 2^n-gon does vary slightly as it slides around the ellipse.

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