I've been playing with inscribing an equilateral triangle inside the standard ellipse (x/a)^2+(y/b)^2=1. In order to keep Macsyma from going nuts, I first pinned down the point (x1,y1)=(a,0). I then eliminated y3,x3,y2, leaving only an equation in x2. Alternatively, we can eliminate y3,x3,x2, leaving an equation in y2. I then eliminated the complex solutions & matched up the real x2's and y2's. Due to symmetry, the x3's would be the same as the x2's. Curiously, the equation involves factors like (a-b) and (a+b), but also sqrt(3a^2+b^2), and most curiously sqrt(3b^2-7a^2). There appears to be some sort of interesting behavior that happens when 3b^2=7a^2. Let a=sqrt(3), b=sqrt(7). Then the ellipse equation is 7xx+3yy=21. The equilateral triangle in this case is (sqrt(3),0), (-3sqrt(3)/4,7/4), (-3sqrt(3)/4,-7/4). This particular ratio of a/b=sqrt(3/7) might be interesting, because (**speculation**) it might produce the equilateral triangle of largest area (relative to the area of the ellipse).