An Extra Triple of Equilateral Triangles for Napoleon

Equilateral Triangles in Napoleon's Configuration

Let $ABC',$ $BCA',$ and $CAB'$ be Napoleon's triangles constructed on the sides of $\Delta ABC.$ Define $A_B$ and $A_C$ as the intersections of $A'B$ and, respectively, $A'C$ with the circumcircle $C(ABC)$ of $\Delta ABC.$ Define similarly $B_A,B_C,C_A,C_B.$

The triangles $A_{B}B_{A}C,$ $AB_{C}C_{B},$ and $A_{C}BC_{A}$ are equilateral.

Hint

The problem is very simple; it submits to chasing inscribed angles.

Solution

Consider, for example, $\Delta AB_{C}C_{B}$ and, more specifically, its side $AB_{C}.$ This is a chord in $C(ABC).$

Since $\Delta CAB'$ is equilateral $\angle ACB_C=120^{\circ},$ meaning that side $AB_{C}$ subtends one third of the circumcircle. The same holds for the other sides and for the remaining two triangles.

Acknowledgment

This problem has been posted by Dao Thanh Oai at the CutTheKnotMath facebook page.