Question

面积为 $2002$ 的凸四边形 $ABCD$ 内有一点 $P$ ，满足 $PA=24$ 、 $PB=32$ 、 $PC=28$ 、且 $PD=45$ 。求 $ABCD$ 的周长。

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 $ABCD.$

$4\sqrt{2002}$

$2\sqrt{8465}$

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

$2\sqrt{8633}$

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

答案：E

知识点：面积对角线勾股定理极限情形界定

难度评级：2150

解答：

任意四边形的面积不超过 $\tfrac12\,d_1 d_2$ ，其中 $d_1,d_2$ 为对角线长；等号恰在对角线垂直时成立。这里 $\begin{gathered} 2002=\text{Area} \\ {}\le \tfrac12\,AC\cdot BD \\ {}\le \tfrac12(PA+PC)(PB+PD) \\ {}= \tfrac12\cdot52\cdot77 \\ {}= 2002. \end{gathered}$

因此等号成立，两条对角线垂直并在 $P$ 相交。于是 $\begin{aligned} AB &= \sqrt{24^2+32^2}=40, \\ BC &= \sqrt{28^2+32^2}=4\sqrt{113}, \end{aligned}$ $\begin{aligned} CD &= \sqrt{28^2+45^2}=53, \\ DA &= \sqrt{45^2+24^2}=51. \end{aligned}$

周长为 $144+4\sqrt{113}=4\left(36+\sqrt{113}\right)$ 。

所以正确答案是 E。

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.

2002 AMC 12B 第 24 题

答案：E

难度评级：2150

解答：

其他年份的第 24 题