2002 AMC 12B Problem 24

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)

For a quadrilateral with diagonals

d_1,d_2,

the area is at most

\tfrac12 d_1 d_2,

with equality when the diagonals are perpendicular

Here

\tfrac12(PA+PC)(PB+PD)

=2002

forces perpendicular diagonals meeting at

P

; find each side with the Pythagorean theorem

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

\begin{gathered} 2002=\text{Area} \\ {}\le \tfrac12\,AC\cdot BD \\ {}\le \tfrac12(PA+PC)(PB+PD) \\ {}= \tfrac12\cdot52\cdot77 \\ {}= 2002. \end{gathered}

Equality forces the diagonals to be perpendicular and to intersect at

P.

Then

\begin{aligned} AB &= \sqrt{24^2+32^2}=40, \\ BC &= \sqrt{28^2+32^2}=4\sqrt{113}, \end{aligned}

\begin{aligned} CD &= \sqrt{28^2+45^2}=53, \\ DA &= \sqrt{45^2+24^2}=51. \end{aligned}

The perimeter is

144+4\sqrt{113}=4\left(36+\sqrt{113}\right).

Thus, the correct answer is E.

Problem 24 in other years