Let $ABC$ be a fixed triangle in the plane. Let $D$ be an arbitrary point in the plane. The circle with center $D$, passing through $A$, meets $AB$ and $AC$ again at points $A_b$ and $A_c$ respectively. Points $B_a, B_c, C_a$ and $C_b$ are defined similarly. A point $D$ is called good if the points $A_b, A_c,B_a, B_c, C_a$, and $C_b$ are concyclic. For a given triangle $ABC$, how many good points can there be? (A. Garkavyj, A. Sokolov )
Problem
Source: 2014 Sharygin Geometry Olympiad Final Round 10.4
Tags: geometry, Concyclic
Pathological
03.05.2019 19:49
Cool problem! We'll claim that there are always exactly $\boxed{4}$ good points. First of all, notice that the circumcenter $O$ of $\triangle ABC$ is clearly always a good point, so we'll just show that there are exactly $3$ other good points $D \neq O$. Suppose that $D \neq O$ is some good point. Let $E$ be the isogonal conjugate of $D$ w.r.t $\triangle ABC$ (could be a point at infinity). Since the circumcenter of the circle through $A_b, A_c, B_a, B_c, C_a, C_b$ must lie on the perpendicular bisectors of $A_bB_a, B_cC_b, C_aA_c$, we know that it must be the reflection of $O$ over $D,$ call this point $D'.$ Therefore, by $D'A_b = D'A_c$ and $DA_b = DA_c = DA,$ we know that $DD' \perp A_bA_c$, i.e., $OD \perp A_bA_c.$ This clearly implies that $AE || OD$ (or $E = A$), with similar conditions for the other two vertices. If $E$ is one of the vertices, say $E = A$ WLOG, then it's easy to see that $BE || OD$ and $CE || OD$ cannot simultaneously hold for the simple reason that $BA$ and $CA$ aren't parallel. Henceforth, let's assume that $E$ isn't one of the vertices. Since $AE || BE || CE$, we know that $E$ must be the point at infinity along line $AE$. This is known to be equivalent to $D$ lying on $(\triangle ABC),$ which follows just by some simple angle-chasing. Therefore, it suffices now to find all $D \in (\triangle ABC)$ so that $OD$ is parallel to the reflection of $AD$ in the angle-bisector of $\angle BAC.$ We can now easily see that there are exactly three such $D$, e.g. by noting that as we rotate $D$ around the circle $OD$ traverses an angle of $2\pi$ and the other line traverses an angle of $\pi$ in the other direction, and so there'd be $3$ by continuity for example. Adding this back to our other good point ($O$), we have a grand total of $\boxed{4}$ good points, as claimed. $\square$