Solving along at home:

1: Pretty much the same as Davi Medeiros (post 6), except I used instead of .

2: Pretty much the same as SCP (post 2), except I was too lazy to describe the lower-bound peaceful configuration with anything other than a diagram.

3: So I figured out that was one of the intersections of the circle through and centered on , and likewise for . I then followed it up with an obscenely contrived sequence of homotheties, reflections, and inversions that I wouldn't commit to paper for a million dollars.

Fine, that's hyperbole, I'm going to do it now. It looks like this:

Let the aforementioned circles through resp. be resp. . Let those circles' centers be resp. and let those circles' other intersections with resp. be resp. .

Note that and are isogonals with respect to .

We want to prove that the perpendicular bisectors of and intersect on . Observe that and are respectively parallel to those perpendicular bisectors (look at homotheties centered at resp. with ratio 2) and they intersect at , so if we can establish a homothety centered at that sends to and to we're done. This is true iff the subdiagrams and are similar.

Now, compose an inversion centered at with radius with a reflection about the angle bisector of (it doesn't matter in what order, they commute); this sends and to each other, so it sends to . This proves that the two subdiagrams are similar because the angle between the lines of tangency from to the circles is preserved at each step.

Click for diagram, which is way too large, oops

What can I say? Once a transformation addict, always a transformation addict?

Amusingly, I didn't use the points and at all or the relation they imply between and , except for implicitly at the very end to prove that the ending condition is equivalent to the problem's condition. So it looks like I proved a generalization of the problem. (!) Or else this is a fakesolve. (!) I hope a reader who is better at geometry than I am can tell me which one it is.