Let $ABC$ be an equilateral triangle. Let $A_1,B_1,C_1$ be interior points of $ABC$ such that $B{A}_1=A_{1}C$, $C{B}_1=B_{1}A$, $A{C}_1=C_{1}B$, and

Let $B{C}_1$ and $C{B}_1$ meet at $A_2$, let $C{A}_1$ and $A{C}_1$ meet at $B_2$, and let $A{B}_1$ and $B{A}_1$ meet at $C_2$. Prove that if triangle $A_1{B}_1{C}_1$ is scalene, then the three circumcircles of triangles $A{A}_2{A}_1$, $B{B}_2{B}_1$, and $C{C}_2{C}_1$ all pass through two common points.

Step 1 (Symmetry): $A_1,B_1,C_1$ are on the perpendicular bisectors of sides $BC,CA,AB$ respectively (since $B{A}_1=A_{1}C$, etc.). The angle condition $\angle B{A}_{1}C+\angle C{B}_{1}A+\angle A{C}_{1}B=480^{\circ }$ constrains how 'deep' these points are.

Step 2 (Points $A_2,B_2,C_2$): These are intersections of cevians. By Ceva's theorem considerations, the configuration has a natural trilinear symmetry.

Step 3 (Circumcircles): The circumcircle of $A{A}_2{A}_1$ passes through $A$, $A_2$, $A_1$. Similarly for the other two. To show all three share two common points, show the radical axes of any two pairs are the same line, hence all three circles are coaxial.

Step 4 (Radical axis computation): The radical axis of the circumcircles of $A{A}_2{A}_1$ and $B{B}_2{B}_1$ passes through the radical center of these with the circumcircle of $C{C}_2{C}_1$. The $480^{\circ }$ condition and equilateral triangle symmetry force a common radical axis, hence two common points.

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