ABC is equilateral iff the midpoints of the sides and of the segments from vertix to centroid all lie on a circumference

Question

I have come across a geometry problem that states:

Let $ABC$ be an acute triangle. Let $AM$, $BN$ and $CL$ be the medians, which intersect in the centroid $G$. Let $M'$, $N'$ and $L'$ be the midpoints of $AG$, $BG$ and $CG$, respectively. Show that the six points $M$, $M'$, $N$, $N'$, $L$, $L'$ all lie on a circumference if and only if $ABC$ is equilateral.

I thought that a different proof from the canonical one might involve shear mapping, meaning that moving one of the three vertices, one could prove that the whole triangle transforms, non rigidly, and the points intersect on a conical, but not a circumference anymore.

I don't know much of this, I have found the related article on Wikipedia but wouldn't know how to prove anything with it. Can anyone help?

Answer 1

HINT.-In the attached figure there is an equilateral triangle with vertices A,B,C and the corresponding points M,N,L,M',N',L', calculated by their coordinates as well as the circumference of the case. It is verified that the last six points are indeed cocyclical. To prove that said six points are not on the same circle if the triangle is not equilateral, we can leave the same side AB and only vary the position of vertex C, which will make the corresponding points not cocyclic. But this calculation will be somewhat more tedious than the one presented. However, the circumference is determined by the points M,N,L and it would be enough to verify that one of the points M',N',L' is not on this circumference. This would be an analytical proof and perhaps a properly geometric one is shorter, but the problem does not have an immediate solution.

Answer 2

If $\triangle ABC$ is equilateral, then it's very easy to show that $NLML'$ is cyclic. Similarly, $NMLM'$ and $LNMN'$ are cyclic. Hence $M,N,L,M',N',L'$ lie on one circle.

Conversely, let $C$ be the circle passing through $M,N,$ and $L$. Note that $M',N',$ and $L'$ also lie on $C$.

Since $M'$ and $N'$ are the midpoints, we have $M'N'||AB||MN.$ Hence, $$\angle GM'N'=\angle GMN.$$

However, $\angle GN'M'=\angle GMN$ because $M,N,N'$ and $M'$ lie on the same circle.

Therefore, $\angle GM'N'=\angle GN'M'$, which implies $GM'=GN'$. Consequently, $AG=BG.$ Similarly, we can prove $AG=CG$, which implies $\triangle ABC$ must be equilateral (please note that the three medians are equal, indeed).

We are done.