AMC12 2002 B

AMC12 2002 B · Q24

AMC12 2002 B · Q24. It mainly tests Circle theorems, Coordinate geometry.

A convex 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 $ABCD$.

一个面积为 $2002$ 的凸四边形 $ABCD$ 内部有一点 $P$，使得 $PA = 24, PB = 32, PC = 28, PD = 45$。求 $ABCD$ 的周长。

(A) 4\sqrt{2002} 4\sqrt{2002}

(B) 2\sqrt{8465} 2\sqrt{8465}

(D) 2\sqrt{8633} 2\sqrt{8633}

(E) 4(36 + \sqrt{113}) 4(36 + \sqrt{113})

Answer

Correct choice: (E)

正确答案：（E）

Solution

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, \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, \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 $AB + BC + CD + DA = 4(36 + \sqrt{113}) \Rightarrow \mathrm{(E)}$.

有 \[[ABCD] = 2002 \le \frac 12 (AC \cdot BD)\] （这对任意凸四边形都成立：沿 $AC$ 将四边形分割，再用三角形面积公式分别计算 $[ACB]$ 与 $[ACD]$），且只有当 $\overline{AC} \perp \overline{BD}$ 时取等号。由三角不等式， \begin{align*}AC &\le PA + PC = 52\\ BD &\le PB + PD = 77\end{align*} 当且仅当 $P$ 分别在 $\overline{AC}$ 与 $\overline{BD}$ 上时取等号。因此 \[2002 \le \frac{1}{2} AC \cdot BD \le \frac 12 \cdot 52 \cdot 77 = 2002\] 由于取到了等号，$\overline{AC} \perp \overline{BD}$ 且在点 $P$ 处相交，如下图所示。由勾股定理， \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*} $ABCD$ 的周长为 $AB + BC + CD + DA = 4(36 + \sqrt{113}) \Rightarrow \mathrm{(E)}$。

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