2*A(Pn)= pi- A(Pn-1);
2*A(Pn+1)= pi- A(Pn); eliminating pi we get the difference equation

A(Pn+1)=(A(Pn)+A(Pn-1))/2

which solution render P(k)= (P(0)+P(1)+P(2))/3 and since this sum amount 180 degree P(n)= 60 degrees.

The points P, P1, P2 will be located in a kind of spiral being its limit the circumcicle containing the equilateral triangle circumscribed to the given circle. The trajectory it will depend obviously of the initial selection of P

Each triangle is isoceles by construction, with the current angle A(P_n) between the equal sides. Therefore, 2*A(P_n+1) + A(P_n) = pi. This can be rearranged into the symmetrical form (pi/3 - A(P_n)) / 2 = -(pi/3 - A(P_n+1)). This shows clearly that the angles will alternate between being greater than pi/3 and less than pi/3. Also, at each step, the difference from pi/3 will decrease by a factor of 2. Therefore, the sequence of angles goes exponentially to a stable limit pi/3. Thus the triangles approach being equilateral in the limit.

If the initial angle is expressed as pi/3 + A, then the angles are pi/3 + A*(-1/2)^n, where the initial angle is given number n=0. The angle between the ray OP_0 and OP_1 is then pi - 1/2*(pi/3 + A + pi/3 - A/2) = 2pi/3 - A/4. For each succeeding pair of rays, the angle will be 2pi/3 + (A/2)*(-1/2)^n, where the initial angle is given number n=1. This shows that the points P_0, P_1, ... rotate around the center of the circle in unequal steps that converge to an angle of 2pi/3. (This is as it'should be, because the limiting triangle is isoceles, and so the rotation should be 2pi/3 in the limit.) The total angle of rotation between OP_0 and OP_n is then n*2pi/3 - (A/2)*(1/2)^n.

The distances OP_n will then be R*csc(1/2*(pi/3 + A*(-1/2)^n)), where R is the radius of the circle. These distances will alternately be larger or smaller than 2R, and so they will lie on two uneven spirals, one spiral of decreasing radial distances, and another spiral of increasing radial distances. I think that because angles of 2pi can be discarded, the points will lie on 3 nice smooth spirals going inward and 3 nice smooth spirals going outward. This is because taking n = 1, 4, 7, etc makes angle n*2pi/3 equivalent to 2pi/3, and similarly form n = 2, 5, 8, etc and n = 3, 6, 9, etc.

This is as much as I can see right now. I hope it is of interest.

--Stuart Anderson