I've been continuing to work on this problem, if only to sharpen my Macsyma skills. ellipse: (x/a)^2+(y/b)^2=1 The equilateral triangle inscribed in a general ellipse can be handled by Macsyma. If a=1, then the length of the side does depend upon b and upon the location of the first of the points. If a=1, b=1/sqrt(3), then the equilateral triangle goes through the major & minor axes, and is one half of the inscribed quadrilateral for this a,b. The inscribed pentagon has a regular solution, in which the sides don't cross one another, and a degenerate solution in which they do cross (when a/=b). Using ascii art, the two cases look something like these: /\ / \ | | |__| /\ /\ /__\__/__\ \/ For example, if a=1, b=2, the first case is approximately like (1,0), (0.5798,+-1.6296), (-.9072,+-.8414), with side length 1.683. the second case is approximately like (1,0), (-.9926,+-.2428), (0.865,+-1.004), with side length 2.007. This answers the question whether the crossing case has smaller perimeter in the negative. BTW, it is easier in Macsyma to solve for the y coordinates first, and then the x coordinates. I don't know how to set up a set of polynomial equations in which the degenerate solutions don't occur. Also, when solving the equations for general a,b, you effectively get the solutions for _all conics_, since I don't know how to eliminate the solutions in which a, b are complex numbers.