Proof that the triangle formed by mass centers is equilateral. Let $ABC$ be a triangle and consider $A_1$, $B_1$, $C_1$ outside the triangle such that triangles $ABC_1$, $BCA_1$ and $ACB_1$ are equilaterals.
Consider now $A_2$, $B_2$ and $C_2$ mass centres of $BCA_1$, $ACB_1$ and $ABC1$, respectively. Prove that triangle $A_2B_2C_2$ is a equilateral triangle.
 A: I am working with $A,B,C$ are oriented clockwise, so you can make a sketch. We have $A_2=(B+C+A_1)/3$, $B_2=(A+B+C_1)/3$, $C_2=(A+C+B_1)/3$.
We want to show that if we rotate $\overline{B_2C_2}$  an angle of 60 anti-clockwise we get $\overline{B_2A_2}$. In other words want to show that $R(C_2-B_2)=A_2-B_2$, where $R$ denotes the operation rotation by 60 degrees(wrt origin), note this is a linear map .
Let's compute \begin{align}R(C_2-B_2)&=R(\frac{C+B_1-B-C_1}{3})=R(\frac{C-B+B_1-A+A-C_1}{3})\\
&=\frac{R(C-B)}{3}+\frac{R(B_1-A)}{3}+\frac{R(A-C_1)}{3}
\end{align}
Now, note that $R(C-B)=A_1-B$,  $R(B_1-A)=C-A$ and $R(A-C_1)=B-C_1$ because $BCA_1$, $ACB_1$ and $ABC_1$ are equilateral.
We get 
\begin{align}R(C_2-B_2)=\frac{A_1-B}{3}+\frac{C-A}{3}+\frac{B-C_1}{3}=\frac{B+C+A_1}{3}-\frac{A+B+C_1}{3}=A_2-B_2.
\end{align}
This is what we wanted to prove.
A: Consider the compositions of rotations $I= R_{A_2,120^\circ}\circ R_{B_2,120^\circ}$; note $I(A)=B$. Let $S$ be the point such that oriented angle $\angle(SB_2,B_2A_2)=60^\circ$ and oriented angle $\angle (B_2A_2,SA_2)=60^\circ$. We have:
$$I=R_{A_2,120^\circ}\circ R_{B_2,120^\circ}= S_{SA_2}\circ S_{A_2B_2}\circ S_{A_2B_2}\circ S_{SB_2}= S_{SA_2}\circ S_{SB_2}= R_{S,240^\circ}.$$
Therefore, $R_{S,240^\circ}(A)=B$, so $SA=SB$ and oriented angle $\angle(SA,SB)=240^\circ$. Now we conclude that $S=C_2$. So $\angle(C_2B_2,B_2A_2)=60^\circ$ and $\angle (B_2A_2,C_2A_2)=60^\circ$, hence $\triangle A_2B_2C_2$ is equilateral.
