Let $A$, $B$, $C$ be the vertices of $T$, and let $G$ be its centroid. Let $M_a$, $M_b$, and $M_c$ be the midpoints of the sidelines $BC$, $CA$, and $AB$, respectively. It turns out that the circumcenters of the six triangles $AGM_c$, $BGM_c$, $ BGM_a$, $CGM_a$, $CGM_b$, and $AGM_b$ lie on a common circle, which is the van Lamoen circle of $T$.
$M_a,M_b,M_c$分别为$BC,CA,AB$的中点,则六个三角形$AGM_c$, $BGM_c$, $ BGM_a$, $CGM_a$, $CGM_b$ 和 $AGM_b$的外心共圆.
math.hkust.edu.hk/excalibur/v6_n1.pdf#page=2
For the second problem, as the 6 circumcenters of the smaller triangles are not on any circles that we can see immediately, so we may try to use the converse of the intersecting chord theorem. For a triangle $A B C$, let $G, D, E$, $F$ be the centroid, the midpoints of sides $B C, C A, A B$, respectively. Let $O_{1}, O_{2}$, $O_{3}, O_{4}, O_{5}, O_{6}$ be the circumcenters of triangles $D B G, B F G, F A G, A E G, E C G$, $C D G$, respectively.
Well, should we draw the 6 circumcircles?
It would make the picture complicated. The circles do not seem to be helpful at this early stage. We give up on drawing the circles, but the circumcenters are important. So at least we should locate them. To locate the circumcenter of $∆FAG$, for example, which two sides do we draw perpendicular bisectors? Sides $AG$ and $FG$ are the choices because they are also the sides of the other small triangles, so we can save some work later.
Trying this out, we discover these perpendicular bisectors produce many parallel lines and parallelograms!
Since circumcenters are on perpendicular bisectors of chords, lines $O_3 O_4$ , $O_6 O_1$ are perpendicular bisectors of $AG$, $GD$, respectively. So they are perpendicular to line $AD$ and are $\frac12 AD$ units apart.
Aiming in showing $O_{1}$, $O_{2}$, $O_{3}$, $O_{4}$ are concyclic by the converse of the intersecting chord theorem, let $K$ be the intersection of lines $O_{1} O_{2}, O_{3} O_{4}$ and $L$ be the intersection of the lines $O_{4} O_{5}, O_{6} O_{1}$. Since the area of the parallelogram $KO_{4} L O_{1}$ is
$$
\frac{1}{2} A D \cdot K O_{4}=\frac{1}{2} B E \cdot K O_{1} \text {, }
$$
we get $K O_{1} / K O_{4}=A D / B E$.Now that we get ratio of $K O_{1}$ and $K O_{4}$, we should examine $KO_{2}$ and $KO_{3}$. Trying to understand $△KO_{2} O_{3}$, we first find its angles. Since $K O_{2} \perp B G$, $O_{2} O_{3} \perp FG$ and $KO_{3} \perp AG$, we see that $\angle K O_{2} O_{3}=\angle B G F$ and $\angle K O_{3} O_{2}=$ $\angle F G A$. Then $\angle O_{2} K O_{3}=\angle D G B$. At this point, you can see the angles of $△KO_{2} O_{3}$ equal the three angles with vertices at $G$ on the left side of segment $A D$.
Now we try to put these three angles together in another way to form another triangle. Let $M$ be the point on line $A G$ such that $M C$ is parallel to $B G$. Since $\angle M C G=\angle B G F, \angle M G C=\angle F G A$ (and $\angle G M C=\angle B G D$, ) we see $
△KO_{2} O_{3}, M CG$ are similar.
The sides of $\triangle M C G$ are easy to compute in term of $A D, B E, C F$. As $A D$ and $B E$ occurred in the ratio of $K O_{1}$ and $K O_{4}$, this is just what we need! Observe that $\triangle M C D, G B D$ are congruent since $\angle M C D=\angle G B D$ (by $M C$ parallel to $G B$ ), $C D=B D$ and $\angle M D C=\angle G D B$. So
$$
M G=2 G D=\frac{2}{3} A D,
M C=G B=\frac{2}{3} B E
$$
(and $C G=\frac{2}{3} C F$. Incidentally, this means the three medians of a triangle can be put together to form a triangle! Actually, this is well-known and was the reason we considered $\triangle M C G$.) We have $KO_{3} / K O_{2}=M G / M C=A D / B E=$ $K O_{1} / K O_{4}$.
So $K O_{1} \cdot K O_{2}=K O_{3} \cdot K O_{4}$, which implies $O_{1}, O_{2}, O_{3}, O_{4}$ are concyclic. Similarly, we see that $O_{2}, O_{3}, O_{4}, O_{5}$ concyclic (using the parallelogram formed by the lines $O_{1} O_{2}, O_{4} O_{5}, O_{2} O_{3}, O_{5} O_{6}$ instead) and $O_{3}, O_{4}, O_{5}, O_{6}$ are concyclic.
