A [convex](https://artofproblemsolving.com/wiki/index.php/Convex_polygon) [quadrilateral](https://artofproblemsolving.com/wiki/index.php/Quadrilateral) $ABCD$ with area $2002$ contains a point $P$ in its interior such that $PA = 24, PB = 32, PC = 28, PD = 45$. Find the perimeter of $ABCD$.
$\mathrm{(A)}\ 4\sqrt{2002} \qquad \mathrm{(B)}\ 2\sqrt{8465} \qquad \mathrm{(C)}\ 2$ $(48+\sqrt{2002}) \qquad \mathrm{(D)}\ 2\sqrt{8633} \qquad \mathrm{(E)}\ 4(36 + \sqrt{113})$
We have 
\[[ABCD] = 2002 \le \frac 12 (AC \cdot BD)\]
(This is true for any convex quadrilateral: split the quadrilateral along $AC$ and then using the triangle area formula to evaluate $[ACB]$ and $[ACD]$), with equality only if $\overline{AC} \perp \overline{BD}$. By the [triangle inequality](https://artofproblemsolving.com/wiki/index.php/Triangle_inequality),  
\begin{align*}AC &\le PA + PC = 52\\ BD &\le PB + PD = 77\end{align*}
with equality if $P$ lies on $\overline{AC}$ and $\overline{BD}$ respectively. Thus
\[2002 \le \frac{1}{2} AC \cdot BD \le \frac 12 \cdot 52 \cdot 77 = 2002\]
Since we have the equality case, $\overline{AC} \perp \overline{BD}$ at point $P$, as shown below. 


By the [Pythagorean Theorem](https://artofproblemsolving.com/wiki/index.php/Pythagorean_Theorem),  
\begin{align*} AB = \sqrt{PA^2 + PB^2} & = \sqrt{24^2 + 32^2} = 40\\ BC = \sqrt{PB^2 + PC^2} & = \sqrt{32^2 + 28^2} = 4\sqrt{113}\\ CD = \sqrt{PC^2 + PD^2} & = \sqrt{28^2 + 45^2} = 53\\ DA = \sqrt{PD^2 + PA^2} & = \sqrt{45^2 + 24^2} = 51 \end{align*}
The perimeter of $ABCD$ is $AB + BC + CD + DA = 4(36 + \sqrt{113}) \Rightarrow \mathrm{(E)}$.