Constructing equilateral triangle from three equilateral triangles. I have heard about this problem some time ago and don't remember all the details.Therefore problem's conditions may have some mistakes.
Problem:

We have three arbitrary (size of edges, location and orientation in space are random) equilateral triangles ABC, DEF and GHI. If we draw lines from corresponding
  corners we create three new triangles. After defining centroids of new triangles and connecting them, we always get another equilateral triangle. Question is to proof this.


I tried drawing and it seemed quite true. Then I try to calculate but It take too much time. After I decided, to get solution using python and sympy. Formulas that I get from them didn't match. But I think there is mistake in program.There is link to notebook code .
 A: I guess, I can help with solution when initial triangles are simultaneously labeled clockwise or counterclockwise.
Consider everything happens in a complex plane, thus every point corresponds to some complex number $z = x + yi$.
Let's consider how an equilateral triangle can be defined within this setup.
I propose the following.
Let a complex number $z$ define its centroid.
Now I define one of its vertices as $z + v$, where $v$ is some other complex number.
This information is enough to fully recover triangle's position and rotation, thus two other vertices may be written in terms of $z$ and $v$ as $z + v e^{2\pi i/3}$ and $z + v e^{-2\pi i/3}$ (I omit the proof as it seems quite obvious).
Now the original problem.
I can write coordinates of vertices in terms of parameters $z_1$, $z_2$, $z_3$, $v_1$, $v_2$, $v_3$ as follows (rows of the table describe the original triangles, while reading through columns we see points in a rearranged order, thus "new triangles" in problem statement)
$$
\begin{array}{|c|c|c|c|}
\hline
  \text{Triangle/Triangle} & \triangle ADG & \triangle BEH & \triangle CFI \\ 
\hline
   \triangle ABC & z_1 + v_1 & z_1 + e^{2\pi i/3} v_1 & z_1 + e^{-2\pi i/3} v_1 \\
\hline
   \triangle DEF & z_2 + v_2 & z_2 + e^{2\pi i/3} v_2 & z_2 + e^{-2\pi i/3} v_2 \\
\hline
   \triangle GHI & z_3 + v_3 & z_3 + e^{2\pi i/3} v_3 & z_3 + e^{-2\pi i/3} v_3 \\
\hline
\end{array}
$$
Now we can write centroids for new triangles
$$
\begin{align}
O &= \frac{z_1 + z_2 + z_3 + v_1 + v_2 + v_3}{3}\\
L &= \frac{z_1 + z_2 + z_3 + e^{2\pi i/3}v_1 + e^{2\pi i/3}v_2 + e^{2\pi i/3}v_3}{3}\\
P &= \frac{z_1 + z_2 + z_3 + e^{-2\pi i/3}v_1 + e^{-2\pi i/3}v_2 + e^{-2\pi i/3}v_3}{3}\\
\end{align}
$$
Obviously, they are vertices of an equilateral triangle.
Why?
Lets define
$$
\begin{align}
Z &= \frac{z_1 + z_2 + z_3}{3}\\
V &= \frac{v_1 + v_2 + v_3}{3}
\end{align}
$$
Than we can write
$$
\begin{align}
O &= Z + V\\
L &= Z + e^{2\pi i/3}V\\
P &= Z + e^{-2\pi i/3}V\\
\end{align}
$$
that corresponds to the way, we used to define equilateral triangles.
