18 Nov
2010
18 Nov
'10
7:04 p.m.
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.