Let $A B C$ be a triangle with $A B=2021, A C=2022$, and $B C=2023$. Compute the minimum value of $A P+2 B P+3 C P$ over all points $P$ in the plane.
Solution 1: The minimizing point is when $P=C$. To prove this, consider placing $P$ at any other point $O \neq C$. Then, by moving $P$ from $O$ to $C$, the expression changes by \[ (A C-A O)+2(B C-B O)+3(C C-C O)<O C+2 O C-3 O C=0 \] by the triangle inequality. Since this is negative, $P=C$ must be the optimal point. The answer is $2022+2 \cdot 2023+3 \cdot 0=6068$. Solution 2: We use a physical interpretation. Imagine an object acted upon by forces of magnitudes 1,2 , and 3 towards $A, B$, and $C$, respectively. The potential energy of the object at point $P$ in this system is $A P+2 B P+3 C P$. This potential energy is minimized when the object experiences 0 net force; in this case, it occurs when it is exactly at point $C$ (because the pull towards $C$ overpowers the other two forces combined). $\fbox{6068}$.