Question

A convex quadrilateral $ABCD$ with area $2002$ contains a point $P$ in its interior such that $PA=24,$ $PB=32,$ $PC=28,$ and $PD=45.$ Find the perimeter of $ABCD.$

$4\sqrt{2002}$

$2\sqrt{8465}$

$2\left(48+\sqrt{2002}\right)$

$2\sqrt{8633}$

$4\left(36+\sqrt{113}\right)$

Solution:

For any quadrilateral, the area is at most $\tfrac12\,d_1 d_2$ where $d_1,d_2$ are the diagonals, with equality exactly when they are perpendicular. Here $2002=\text{Area}\le\tfrac12\,AC\cdot BD\le\tfrac12(PA+PC)(PB+PD)=\tfrac12\cdot52\cdot77=2002.$

Equality forces the diagonals to be perpendicular and to intersect at $P.$ Then $AB=\sqrt{24^2+32^2}=40,\quad BC=\sqrt{28^2+32^2}=4\sqrt{113},$ $CD=\sqrt{28^2+45^2}=53,\quad DA=\sqrt{45^2+24^2}=51.$

The perimeter is $144+4\sqrt{113}=4\left(36+\sqrt{113}\right).$

Thus, the correct answer is E.

2002 AMC 12B Problem 24

Solution:

Problem 24 in Other Years