Re: [math-fun] equilateral elliptical n-gons
[My statement below about "notchy" is *wrong*. 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.] I don't see any way to put an equilateral quadrilateral inside an ellipse, except for the diamond and the square. So the rotation thing is definitely "notchy". At 10:55 AM 11/18/2010, James Buddenhagen wrote:
I think it may be more complicated than you describe below, but it will still turn out to be algebraic, in fact if you have one vertex of the equilateral polygon at an ellipse vertex the coordinates of the polygon vertices will be algebraic for any algebraic a,b and any n, because you have to solve a system of polynomial equations (arising from the distance formula and the equation of the ellipse, and forcing closure).
Also, I think the solution will be isolated, no slipping of the vertices around the ellipse as Dan mentioned, unless it is a circle, but I would have to look more closely to be sure
On Thu, Nov 18, 2010 at 12:12 PM, Henry Baker <hbaker1@pipeline.com> wrote:
After a deep discussion with Macsyma, M convinced me that this recursive procedure works, and does indeed produce points whose coordinates are the solutions of polynomials with coefficients in the rational field extended by a & b. Furthermore, these polynomials are Galois-solvable by towers of square roots.
Computing these points is really no more difficult than computing the coordinates of the 2^n-th roots of unity, except that a & b keep entering into the computation. If a=b=1, then the computations are exactly the same as those for calculating the coordinates of the 2^n-th roots of unity.
Basically, given 2 points already on the ellipse, compute the equation of the perpendicular bisector line through those two points. This line intersects the ellipse in two places, _both_ of whom become new additional points (by symmetry). The equation set can be produced by resultant(ellipseeqn,bisectoreqn,y) [a quadratic in x] for the two x coordinates, and resultant(ellipseeqn,bisectoreqn,x) [a quadratic in y] for the two y coordinates. A slightly different approach can produce the points w/o any ambiguity about which x coordinate goes with which y coordinate.
What I meant when I talked about "solutions being isolated" (possibly not cc'ed to math-fun) was that if you have and equilateral n-gon inscribed in an ellipse, it cannot rotate (i.e. points move or slip along the ellipse) without changing the edge lengths, assuming the ellipse does not change, and is not a circle. On Thu, Nov 18, 2010 at 4:10 PM, Henry Baker <hbaker1@pipeline.com> wrote:
[My statement below about "notchy" is *wrong*. 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.]
I don't see any way to put an equilateral quadrilateral inside an ellipse, except for the diamond and the square. So the rotation thing is definitely "notchy".
At 10:55 AM 11/18/2010, James Buddenhagen wrote:
I think it may be more complicated than you describe below, but it will still turn out to be algebraic, in fact if you have one vertex of the equilateral polygon at an ellipse vertex the coordinates of the polygon vertices will be algebraic for any algebraic a,b and any n, because you have to solve a system of polynomial equations (arising from the distance formula and the equation of the ellipse, and forcing closure).
Also, I think the solution will be isolated, no slipping of the vertices around the ellipse as Dan mentioned, unless it is a circle, but I would have to look more closely to be sure
On Thu, Nov 18, 2010 at 12:12 PM, Henry Baker <hbaker1@pipeline.com> wrote:
After a deep discussion with Macsyma, M convinced me that this recursive procedure works, and does indeed produce points whose coordinates are the solutions of polynomials with coefficients in the rational field extended by a & b. Furthermore, these polynomials are Galois-solvable by towers of square roots.
Computing these points is really no more difficult than computing the coordinates of the 2^n-th roots of unity, except that a & b keep entering into the computation. If a=b=1, then the computations are exactly the same as those for calculating the coordinates of the 2^n-th roots of unity.
Basically, given 2 points already on the ellipse, compute the equation of the perpendicular bisector line through those two points. This line intersects the ellipse in two places, _both_ of whom become new additional points (by symmetry). The equation set can be produced by resultant(ellipseeqn,bisectoreqn,y) [a quadratic in x] for the two x coordinates, and resultant(ellipseeqn,bisectoreqn,x) [a quadratic in y] for the two y coordinates. A slightly different approach can produce the points w/o any ambiguity about which x coordinate goes with which y coordinate.
On Thu, Nov 18, 2010 at 5:10 PM, Henry Baker <hbaker1@pipeline.com> wrote:
[My statement below about "notchy" is *wrong*. Given an ellipse, you can start my recursive perpendicular bisector subdivision from _any_ line through the center of the ellipse.
But as someone pointed out, the recursive perpendicular bisector subdivsion doesn't work. You replace side AB with sides AC and CB, where |AC| = |CB|, and you replace side XY with sides XZ and ZY, with |XZ| = |ZY|, but there's no reason to expect that just because |AB| = |XY|, that |AC| should equal |XZ|. Andy
participants (3)
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Andy Latto -
Henry Baker -
James Buddenhagen