Perhaps "sliding" may not be the best word here. The situation is most dramatic with 2,3,4 sided n-gons. In the case of the 2-gon, the length of the sides varies between 2a and 2b, where a,b are the major,minor radii. Similarly, with a rhombus, the length of a side appears to be greatest [sqrt(a^2+b^2)] when it makes a diamond formation, and smallest when it has to be an upright square within the ellipse. As the number of sides increases, the difference in length between a side and the segment of the ellipse it cuts off becomes vanishingly small, so the variation in length during sliding also becomes vanishingly small. --- I still don't know how to construct a non-power of 2 number of vertices -- e.g., an equilateral pentagon -- within an ellipse. At 03:50 PM 11/18/2010, Dan Asimov wrote:
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.