For $ABC$ triangle with $H$ place in the centre of $[AB]$ and $[HC] = 5$
I want to calculate the $xy$ coordinate of $ABC$ from $G$ point centroids of $ABC$ triangle and for example a rotation of $60°$ of $[HC]$ in relation to the x-axis.
I have found the formula to determine the centroids from ABC point: $$Gx=\cfrac{Ax+Bx+Cx}{3 }\\ Gx=\cfrac{Ay+By+Cy}{3}$$ But I can't get it to reverse and add rotation parameters.
Since a circle of centroid $G\left(x_G,\,y_G\right)$ and radius $R > 0$ can be parameterized as:
$$ \left(x,\,y\right) = \left(x_G,\,y_G\right) + R\left(\cos\theta,\,\sin\theta\right) \quad \quad \text{with} \; \theta \in [0,\,2\pi) $$
the vertices of an $n$-sided regular polygon of centroid $G\left(x_G,\,y_G\right)$ and side $L > 0$ are:
$$ \left(x_k,\,y_k\right) = \left(x_G,\,y_G\right) + \frac{L}{2\sin(\pi/n)}\left(\cos\theta_k,\,\sin\theta_k\right) \quad \quad \text{with} \; \theta \in [0,\,2\pi) $$
where $\theta_k = \theta + 2\,k\,\pi/n$, with $k = 0,\,1,\,\dots,\,n-1$.
You should now have all the credentials to adapt it to your needs, good luck! ^_^