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2002 AMC 12B 第 24 题

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所有题目均经美国数学协会（MAA）官方合法授权使用。

24.

面积为 20022002 的凸四边形 ABCDABCD 内有一点 PP，满足 PA=24PA=24、PB=32PB=32、PC=28PC=28、且 PD=45PD=45。求 ABCDABCD 的周长。

A convex quadrilateral ABCDABCD with area 20022002 contains a point PP in its interior such that PA=24,PA=24, PB=32,PB=32, PC=28,PC=28, and PD=45.PD=45. Find the perimeter of ABCD.ABCD.

420024\sqrt{2002}

284652\sqrt{8465}

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

286332\sqrt{8633}

4(36+113)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
小提示：

对角线长为 d1,d2d_1,d_2 的四边形，面积最多为 12d1d2\tfrac12 d_1 d_2，等号在对角线垂直时成立。

For a quadrilateral with diagonals d1,d2,d_1,d_2, the area is at most 12d1d2,\tfrac12 d_1 d_2, with equality when the diagonals are perpendicular

大提示：

此处 12(PA+PC)(PB+PD)\tfrac12(PA+PC)(PB+PD) =2002=2002，迫使对角线垂直并在 PP 相交；用勾股定理求各边。

Here 12(PA+PC)(PB+PD)\tfrac12(PA+PC)(PB+PD) =2002=2002 forces perpendicular diagonals meeting at PP; find each side with the Pythagorean theorem

解答：

任意四边形的面积不超过 12 d1d2\tfrac12\,d_1 d_2，其中 d1,d2d_1,d_2 为对角线长；等号恰在对角线垂直时成立。这里 2002=Area≤12 AC⋅BD≤12(PA+PC)(PB+PD)=12⋅52⋅77=2002。 \begin{gathered} 2002=\text{Area} \\ {}\le \tfrac12\,AC\cdot BD \\ {}\le \tfrac12(PA+PC)(PB+PD) \\ {}= \tfrac12\cdot52\cdot77 \\ {}= 2002 \end{gathered}\text{。}

因此等号成立，两条对角线垂直并在 PP 相交。于是 AB=242+322=40,BC=282+322=4113， \begin{aligned} AB &= \sqrt{24^2+32^2}=40, \\ BC &= \sqrt{28^2+32^2}=4\sqrt{113} \end{aligned}\text{，}CD=282+452=53,DA=452+242=51。 \begin{aligned} CD &= \sqrt{28^2+45^2}=53, \\ DA &= \sqrt{45^2+24^2}=51 \end{aligned}\text{。}

周长为 144+4113=4(36+113)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 12 d1d2\tfrac12\,d_1 d_2 where d1,d2d_1,d_2 are the diagonals, with equality exactly when they are perpendicular. Here 2002=Area≤12 AC⋅BD≤12(PA+PC)(PB+PD)=12⋅52⋅77=2002. \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.P. Then AB=242+322=40,BC=282+322=4113, \begin{aligned} AB &= \sqrt{24^2+32^2}=40, \\ BC &= \sqrt{28^2+32^2}=4\sqrt{113}, \end{aligned} CD=282+452=53,DA=452+242=51. \begin{aligned} CD &= \sqrt{28^2+45^2}=53, \\ DA &= \sqrt{45^2+24^2}=51. \end{aligned}

The perimeter is 144+4113=4(36+113).144+4\sqrt{113}=4\left(36+\sqrt{113}\right).

Thus, the correct answer is E.

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