Geometrically invalid

If $A,B,C$ have position vectors $\mathbf{a},\mathbf{b},\mathbf{c}$ with respect to the circumcentre $O$ as origin (so that $\mathbf{a},\mathbf{b},\mathbf{c}$ all have the same modulus $R$ , the circumradius), then the position vectors of $G$ and $M$ (the midpoint of $BC$ are $\mathbf{g} \; = \; \tfrac13(\mathbf{a}+\mathbf{b}+\mathbf{c}) \hspace{1cm} \mathbf{m} = \tfrac12(\mathbf{b}+\mathbf{c})$ and we require that $\mathbf{g}\cdot(\mathbf{a}-\mathbf{m}) = 0$ . Thus $\begin{array}{rcl} 0 & = & (\mathbf{a}+\mathbf{b}+\mathbf{c}) \cdot (2\mathbf{a} - \mathbf{b} - \mathbf{c}) \; = \; 2|\mathbf{a}|^2 + \mathbf{a}\cdot(\mathbf{b}+\mathbf{c}) - |\mathbf{b}+\mathbf{c}|^2 \\ & = & \mathbf{a}\cdot\mathbf{b} + \mathbf{a}\cdot\mathbf{c} - 2 \mathbf{b}\cdot\mathbf{c} \\ & = & \tfrac12\big[2R^2 - AB^2\big] + \tfrac12\big[2R^2 - AC^2\big] - \big[2R^2 - BC^2\big] \end{array}$ and hence $BC^2 = \tfrac12(AB^2 + AC^2)$ . In our case, we deduce that $BC^2 = \boxed{325}$ .