# 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.

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.
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.