A convex quadrilateral $ABCD$ with area $168$ contains a point $P$ in its interior such that $PA=9, PB=12, PC=5, PD=12$. Find the perimeter of $ABCD$.

1. $[ABCD]\le\frac12(PA\cdot PB+PB\cdot PC+PC\cdot PD+PD\cdot PA)=\frac12(108+60+60+108)=168$, with equality forcing all four angles at $P$ to be right angles.
2. Then $AB=\sqrt{9^2+12^2}=15$, $BC=\sqrt{12^2+5^2}=13$, $CD=\sqrt{5^2+12^2}=13$, $DA=\sqrt{12^2+9^2}=15$.
3. Perimeter $=15+13+13+15=\boxed{56}$.