Re: [math-fun] equilateral elliptical n-gons
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.
Equilateral pentagon in ellipse. For one vertex at (a,0) on x^2/a^2 + y^2/b^2 = 1, set up 4 polynomial equations in 4 unknowns, the coordinates, and solve simultaneously. I tried it for a=5, b=3. A picture is here: http://dl.dropbox.com/u/531485/equilateral_pentagon_in_ellipse.gif The x-coordinates are roots of 4-degree polynomials over the integers, which I can provide, but they are pretty ugly. -- On Thu, Nov 18, 2010 at 6:26 PM, Henry Baker <hbaker1@pipeline.com> wrote:
Perhaps "sliding" may not be the best word here.
...snip...
I still don't know how to construct a non-power of 2 number of vertices -- e.g., an equilateral pentagon -- within an ellipse.
On Thu, Nov 18, 2010 at 7:26 PM, Henry Baker <hbaker1@pipeline.com> wrote:
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.
How do you do a power-of-2 number of vertices; the perpendicular bisector construction doesn't work. What do you mean by "construct"? If we are looking for an N-gon that includes the point (x0, y0), let the coordinates of the desired points be (xi, yi), 0 <= i < N. So we have 2N - 2 unknowns. We have N-1 equations of the form (xi - xj) ^2 + (yi - yj)^2 = (xj - xk)^2 + (yj - yk) ^2 where i+2 = j+1 = k, expressing the fact that the sides are equal, and N-1 more equations expressing the fact that (xi, yi) lies on the ellipse. Solve this set of 2N-2 quadratic equations in 2N-2 unknowns, and you have your polygon. I think a set of quadratics like this is always solvable using only square roots and field operations. Andy
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.
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participants (3)
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Andy Latto -
Henry Baker -
James Buddenhagen