archimedes26 wrote:

Let $K$ be the symmedian point of $ABC$ . Let $X,Y,Z$ be the projections of $K$ on perpendicular bisectors of $BC, CA, AB$
* $ABC$ and $XYZ$ have same centroid.

Let $O, G$ be circumcenter, centroid of $\triangle ABC;$ $AK$ intersects $BC, (O)$ at $D, S;$ $M, Q$ be midpoint of $BC, AS;$ $R$ be intersection of tangents at $B, C$ of $(O)$ . Since $\dfrac{\overline{QD}}{\overline{QS}} = \dfrac{\overline{QS}}{\overline{QR}} = \dfrac{\overline{AQ}}{\overline{QR}} = \dfrac{\overline{AQ} + \overline{QD}}{\overline{QR} + \overline{AQ}} = \dfrac{\overline{AD}}{\overline{AR}}$ we have $\dfrac{\overline{GM}}{\overline{GA}} \cdot \dfrac{\overline{QA}}{\overline{QR}} \cdot \dfrac{\overline{XR}}{\overline{XM}} = \dfrac{- 1}{2} \cdot \dfrac{- \overline{QS}}{\overline{QR}} \cdot \dfrac{\overline{KR}}{\overline{KD}} = \dfrac{\overline{QS}}{\overline{QR}} \cdot \dfrac{\overline{AR}}{\overline{AD}} = 1$ Then $X, G, Q$ are collinear. It's easy to prove that $\triangle ABC$ $\stackrel{-}{\sim}$ $\triangle XYZ$ . Suppose that $N, P$ are midpoint of $CA, AB$ . We have $(YZ, YQ) \equiv (OZ, OQ) \equiv (AB, AQ) \equiv (NP, NQ) \pmod \pi$ $(ZQ, ZY) \equiv (OQ, OY) \equiv (AQ, AC) \equiv (PQ, PN) \pmod \pi$ So $\triangle QYZ$ $\stackrel{+}{\sim}$ $\triangle QNP$ . Hence $\dfrac{QY}{QZ} = \dfrac{QN}{QP} = \dfrac{AN}{AP} = \dfrac{CA}{AB} = \dfrac{ZX}{XY}$ or $XQ$ is $X$ - median of $\triangle XYZ$ . But $X, G, Q$ are collinear then $XG$ is $X$ - median of $\triangle XYZ$ . Similarly, we have $G$ is centroid of $\triangle XYZ$