Triangle of centroids

[Hill]

Homework Statement

Draw an arbitrary triangle, and on each side draw an equilateral triangle lying outside the given triangle. What do you suspect is special about the new triangle formed by joining the centroids (cf. Ex. 19) of the equilateral triangles?
Use complex algebra to prove that you are right.
What happens if the equilateral triangles are instead drawn on the inside of the given triangle?

Relevant Equations

from the "Ex.19", centroid g of a triangle abc is g=1/3(a+b+c)

Here is my sketch:

The triangle $abc$ is arbitrary, the triangles $acp$, $abq$, and $bcr$ are equilateral with centroids $m$, $n$, and $k$. I suspect that the triangle $mnk$ is equilateral. Here is my proof.

By the equation for centroids,
$3m=a+c+p$
$3n=a+q+b$
$3k=b+c+r$

Let's show that angle between $\vec {mk}$ and $\vec {mn}$ is $\pi /3$.
$3 \vec {mk} = 3(k-m)=(b+c+r)-(a+c+p)=(b-a)+(r-p)=\vec {ab} + \vec {pr}=\vec {ab}+\vec{pc}+\vec{cr}$
$3 \vec {mn} = 3(n-m)=(a+q+b)-(a+c+p)=(b-c)+(q-p)=\vec {cb} + \vec {pq}=\vec {cb}+\vec{pa}+\vec{aq}$

Let $R$ be rotation by $\pi /3$, i.e., multiplication by $e^{i \pi /3}$. Then
$R(3 \vec {mk})=R(\vec {ab})+R(\vec{pc})+R(\vec{cr})=\vec{aq}+\vec{pa}+\vec {cb}=3 \vec {mn}$
Done.

If the equilateral triangles are instead drawn on the inside of the given triangle, then R should rotate in opposite sense, i.e., should be replaced by $e^{-i \pi /3}$.

My questions:
1. Does this proof "use complex algebra"?
2. Is there a more direct proof?
3. Is there anything else "special about the new triangle"?

 

[Lnewqban]

3. Is there anything else "special about the new triangle"?

It seems to me that the locations of both centroids are coincident.

Please, see:
https://www.mathsisfun.com/definitions/median-of-triangle.html

 

[Hill]

It seems to me that the locations of both centroids are coincident.

Please, see:
https://www.mathsisfun.com/definitions/median-of-triangle.html

Yes, they are.

Triangles $mca$, $nab$, $kbc$ are similar. Thus,
$m-c=(a-c)z$
$n-a=(b-a)z$
$k-b=(c-b)z$

Summing these equations gives
$m+n+k=a+b+c$

Thank you.

 

[Lnewqban]

 

[Hill]

View attachment 344525

I think, I've covered it here:

If the equilateral triangles are instead drawn on the inside of the given triangle, then R should rotate in opposite sense

Nice drawings. What software do you use?

 

[Lnewqban]

I think, I've covered it here:

Nice drawings. What software do you use?

Yes, I just had some spare time and wanted to verify everything drawing the solution in CAD.
Just posting for future visitors of this thread.
Thank you.