Remember the 7-11 problem where Professor O'Blivet buys four items whose sum and product both total 7.11? There is only one integer (cents) solution, but a graphical method on a certain "elliptic" curve yields infinitely many rational solutions. The curve looks like a "rounded triangle" amidst three "hyperbolas" (http://www.ippi.com/rwg/711.gif). Any straight line cuts these four arcs in exactly three places (two, if one is a graze), and all three cuts represent rational solutions if any two of them do. Inventor and curve connoisseur Alan Adler recently sought some rounded triangles, so I offered a b (x - a) (x - b) (x + -----) 2 b + a y = ---------------------------. a b 3 x - ----- + b + a b + a For a<x<b, this gives an "equilateral" version of the 7-11 rounded triangle ("threelipse"?), when a<0<-a<b<-2a. As b (= the circumradius) varies between -a and -2a, the curve morphs between a circle and an equilateral triangle. The up-down symmetry is obvious from the equation. Less so the 120 degree rotational symmetry. But x = r cos(t), y = r sin(t) gives 0 = (b+a)*r^3*cos(3*t)-(b^2+a*b+a^2)*r^2+a^2*b^2, which is unperturbed by t <- t - 2pi/3. Who was the idiot who misappropriated "hyperboloid" to name a surface? --rwg Pedants depants pentads.