AMC12 2006 B

AMC12 2006 B · Q21

AMC12 2006 B · Q21. It mainly tests Quadratic equations, Circle theorems.

Rectangle $ABCD$ has area $2006$. An ellipse with area $2006\pi$ passes through $A$ and $C$ and has foci at $B$ and $D$. What is the perimeter of the rectangle? (The area of an ellipse is $ab\pi$ where $2a$ and $2b$ are the lengths of the axes.)

矩形 $ABCD$ 的面积为 $2006$。一个面积为 $2006\pi$ 的椭圆通过 $A$ 和 $C$，且焦点在 $B$ 和 $D$ 处。该矩形的周长是多少？（椭圆的面积为 $ab\pi$，其中 $2a$ 和 $2b$ 是其轴的长度。）

(A) $\frac{16\sqrt{2006}}{\pi}$ $\frac{16\sqrt{2006}}{\pi}$

(B) $\frac{1003}{4}$ $\frac{1003}{4}$

(D) $6\sqrt{2006}$ $6\sqrt{2006}$

(E) $\frac{32\sqrt{1003}}{\pi}$ $\frac{32\sqrt{1003}}{\pi}$

Answer

Correct choice: (C)

正确答案：（C）

Solution

Let the rectangle have side lengths $l$ and $w$. Let the axis of the ellipse on which the foci lie have length $2a$, and let the other axis have length $2b$. We have \[lw=ab=2006\] From the definition of an ellipse, $l+w=2a\Longrightarrow \frac{l+w}{2}=a$. Also, the diagonal of the rectangle has length $\sqrt{l^2+w^2}$. Comparing the lengths of the axes and the distance from the foci to the center, we have \[a^2=\frac{l^2+w^2}{4}+b^2\Longrightarrow \frac{l^2+2lw+w^2}{4}=\frac{l^2+w^2}{4}+b^2\Longrightarrow \frac{lw}{2}=b^2\Longrightarrow b=\sqrt{1003}\] Since $ab=2006$, we now know $a\sqrt{1003}=2006\Longrightarrow a=2\sqrt{1003}$ and because $a=\frac{l+w}{2}$, or one-fourth of the rectangle's perimeter, we multiply by four to get an answer of $\boxed{8\sqrt{1003}}$.

设矩形的边长为 $l$ 和 $w$。设椭圆上焦点所在的轴长为 $2a$，另一条轴长为 $2b$。则 \[lw=ab=2006\] 由椭圆的定义，$l+w=2a\Longrightarrow \frac{l+w}{2}=a$。另外，矩形的对角线长为 $\sqrt{l^2+w^2}$。比较椭圆两条轴的长度与焦点到中心的距离，有 \[a^2=\frac{l^2+w^2}{4}+b^2\Longrightarrow \frac{l^2+2lw+w^2}{4}=\frac{l^2+w^2}{4}+b^2\Longrightarrow \frac{lw}{2}=b^2\Longrightarrow b=\sqrt{1003}\] 由于 $ab=2006$，可得 $a\sqrt{1003}=2006\Longrightarrow a=2\sqrt{1003}$。又因为 $a=\frac{l+w}{2}$，即矩形周长的四分之一，所以乘以 $4$ 得答案为 $\boxed{8\sqrt{1003}}$。

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