Henry> The proof a non-existence for n=3 needs to be tightened up a little. ----------------------------------------------------------------------------------------- ES>A C1 configuration of circular arcs remains so under inversion about a point, other than possibly extending to infinity. Invert about one of the arc junctions. The two arcs meeting at that junction become parallel rays, one from a finite point P and extending infinitely far to the left, the other from a finite point Q and extending infinitely far to the right. The rays cannot lie on the same line, as then the two arcs would be part of the same circle. If a 3-arc configuration exists, then P and Q are joined by a single circular arc A. The directed tangents to A at P and Q are parallel, so A must undergo an angle change of 2πn. If n=0, A is a line segment, and this is impossible because the rays do not lie on the same line. If n is not zero, A is (one or more) complete circles, and P and Q coincide, again impossible for the same reason. -- Gene This morning Julian sent me: ---------- Forwarded message ---------- From: Julian Ziegler Hunts <julianj.zh@gmail.com> Date: Wed, Mar 14, 2012 at 10:38 AM Subject: Re: [math-fun] circular-arc splines, again To: billgosper@gmail.com Without the "simple" condition, making a C1 curve out of k circular arcs reduces to finding k circles, each tangent to the next (and the last to the first), with an even number of external tangencies (zero if you want to avoid inflections). This turns into the equations |z_i-z_(i+1)|=±r_i±r_(i+1) (same sign for external tangent, different for internal). I can't figure out how to solve this with Mma, and it's too complicated to do by hand except in simple cases. Can Macsyma do better? You can't do three arcs, even if you use two external tangencies. You need at least one internal tangency. Then you need another circle either internally or externally tangent to both of them, which would need to be tangent at the same point (you can get a different one that's internal to one and external to the other, but that gives you a cusp in the resulting curve), leading to a non-simple curve. Julian