How would you calculate the length of one of the sides of an equalateral polygon (of n sides) inscribed within an ellipse ( of any eccentricity ) where all of the vertices exactly touch the perimeter of the ellipse?
I know that when the eccentricity is zero ( i.e a circle ) the formula: r * (sin(180/n) * 2) will suffice. But what about when the eccentricity is greater than zero?
Thank you for your time.
There is no unique answer. For example, when n=3 note that any point of your ellipse is one vertex of an inscribed equilateral triangle. Consequently, there are infinitely many different sizes (when the ellipse is not a circle). For n=4, an inscribed equilateral quadrangle will necessarily be a rhombus. Thus, for any point A on the ellipse, the oppositee end of the diameter through A will be C, while B and D will be opposite ends of the diameter that is perpendicular to AC. Again it is clear that there are infinitely many different sizes for an equilateral 4-gon inscribed in a noncircular ellipse.
Of course, as n goes to infinity the perimeter of the inscribed equilateral n-gon will approach the circumference of the ellipse. But even for that there is no simple formula. See http://mathcentral.uregina.ca/QQ/database/QQ.09.99/proffit1.htmlChris