No We can choose random point $E$ and build triangle $A_2B_2C_2$ such that $C_1,A_1,B_1$ are reflections of $E$ across midpoints of $A_2B_2C_2$

RagvaloD wrote: No We can choose random point $E$ and build triangle $A_2B_2C_2$ such that $C_1,A_1,B_1$ are reflections of $E$ across midpoints of $A_2B_2C_2$ Why is it possible? Please say it more explicitly

A more explicit proof: We claim for any $E$ we can construct such an $ABC$. Note that the midpoints of $EA_1, EB_1, EC_1$ are precisely the midpoints $X, Y, Z$ of $BC, CA, AB$ each. From here, line $BC$ is the line through $X$ parallel to $YZ$ and so on by homothety, so we can indeed explicitly construct such an $ABC$.

Thank you! I noticed that I wrongly translated the problem, but now everything should be correct.

Oh, this is probably much harder then. :shiver: