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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.

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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.

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  • $\begingroup$ +1............... $\endgroup$
    – nonuser
    Commented Apr 2, 2019 at 19:02
  • $\begingroup$ I need a proof for 5th grade. You can't use rotations or this kind of abstract tools $\endgroup$
    – Hector Guy
    Commented Apr 2, 2019 at 19:29
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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.

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  • $\begingroup$ +1.................. $\endgroup$
    – nonuser
    Commented Apr 2, 2019 at 19:02
  • $\begingroup$ I need a 5th grade proof using only elementary geometry $\endgroup$
    – Hector Guy
    Commented Apr 2, 2019 at 19:37
  • $\begingroup$ This is called Napoleon's theorem, I bet you can find geometric proofs if you search for it. But the proof I gave is understandable for anyone as long as you admit that rotation of sum of vectors is a sum of the vectors rotated. Anyway, good luck.In fact, If you draw the vectors I use would be very enlightening. $\endgroup$ Commented Apr 2, 2019 at 19:44

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