Problem 6

By Step 4, $\mathcal C_a,\mathcal C_b,\mathcal C_c$ are the loci $\operatorname{Pow}(P,\omega_b)=k_a\operatorname{Pow}(P,\omega_c)$, $\operatorname{Pow}(P,\omega_c)=k_b\operatorname{Pow}(P,\omega_a)$, and $\operatorname{Pow}(P,\omega_a)=k_c\operatorname{Pow}(P,\omega_b)$. If a point lies on $\mathcal C_a$ and $\mathcal C_b$, multiplying the first two equalities and using $k_ak_bk_c=1$ gives the third equality, so $\mathcal C_a\cap\mathcal C_b\subseteq\mathcal C_c$; cyclically the same holds for every pair. It remains to justify that the intersections are real and distinct. We use the standard crossing-chords criterion: in a strictly convex configuration, if the two marked chords have interlacing endpoints and the corresponding circumcircle arcs lie in opposite sides of each chord, the two circles cross in two distinct real points; equality would mean tangency, while absence of a crossing would put both arcs on the same side. For $\mathcal C_a=(AA_1A_2)$ and $\mathcal C_b=(BB_1B_2)$, the interlacing chords are $A_2A_1$ and $B_2B_1$ in the convex hexagon $A_2,C_1,B_2,A_1,C_2,B_1$. The inequalities $0<\alpha,\beta,\gamma<30^\circ$ put $A$ and $B$ in the opposite sectors determined by these chords, and the scalene hypothesis excludes the equality (tangent) case. Thus $\mathcal C_a$ and $\mathcal C_b$ have two distinct real intersection points $U,V$. The first inclusion puts both $U,V$ on $\mathcal C_c$, proving that the three circumcircles pass through the same two points.