Ig I've seen this problem before but never got around solving it. BMOSL 2008 G4 wrote: A triangle $ABC$ is given with barycentre $G$ and circumcentre $O$. The perpendicular bisectors of $GA, GB$ meet at $C_1$,of $GB,GC$ meet at $A _1$, and $GC,GA$ meet at $B_1$. Prove that $O$ is the barycenter of the triangle $A_1B_1C_1$. Lemma- In a $\triangle ABC$ let $\{P,P^*\}$ be two pairs of Isogonal Conjugates. Then $P^*$ is the Circumcenter of the triangle formed by the reflection of $P$ over $\{BC,CA,AB\}$. This is a well known fact. So not proving it here. Claim:- $G$ is the Symmedian Point of $\triangle A_1B_1C_1$ Proof-. Let $\{E,F\}=\{GB\cap A_1C_1,GC\cap A_1B_1\}$ and $X$ be the midpoint of $BC$. Note that $$\frac{GE}{GF}=\frac{GB}{GC}=\frac{\sin\angle BGX}{\sin\angle CGX}=\frac{\sin\angle A_1B_1C_1}{\sin\angle A_1C_1B_1}=\frac{A_1C_1}{A_1B_1}$$So, $G$ lies on the $A$- Symmedian of $\triangle A_1B_1C_1$, similarly $G$ lies on the $B,C$ Symmedian of $\triangle ABC$ as well. Hence, $G$ is the Symmedian Point of $\triangle A_1B_1C_1$. Now notice that the triangle formed by the reflection of $G$ over $A_1B_1C_1$ is $\triangle ABC$ and it's Circumcenter is $O$. So, by Lemma $O$ is the Isogonal Conjugate of $G$ WRT $\triangle A_1B_1C_1\implies O$ is the centroid of $\triangle A_1B_1C_1$. $\blacksquare$