10. 3 (IMO 42 Prelim) Let $G$ be the centroid of $\triangle ABC$. Determine the position of point $P$ in the plane of $\triangle ABC$ such that $AP \cdot AG + BP \cdot BG + CP \cdot OG$ is minimized, and express this minimum value in terms of the side lengths of $\triangle ABC$.
10.3 The minimum value is obtained when $P$ is the centroid. The minimum value is $\sum A G^{2}=\frac{1}{3}\left(a^{2}+b^{2}+c^{2}\right)$.

Let $S$ be the circle passing through $B, G, C$, and the median $A L$ intersects $S$ at $G, K$. Let $\angle B G K=\theta, \angle O G L=\varphi, \angle B G N=\delta$ (where $L, N$ are the midpoints of $B C, A B$). The distances from $B, C$ to $G L$ are equal, so we have $B G \cdot \sin \theta=O G \cdot \sin \varphi$, and $B K=2 R \sin \theta, B C=2 R \sin \delta$.

Thus, we have
$$
A G: B C=B G: C K=O G: C K.
$$

Let $P$ be any point on the plane. By the generalized Ptolemy's theorem, we have $P K \cdot B C \leqslant B P \cdot C K+C P \cdot B K$, with equality if and only if $P$ is on $S$. From equation (1), we have $P K \cdot A G \leqslant B P \cdot B G+C P \cdot O G$. Adding $A P \cdot A G$ to both sides, and using $A K \leqslant A P+P K$, we obtain the desired result.