AMC12 2007 B

AMC12 2007 B · Q25

AMC12 2007 B · Q25. It mainly tests Coordinate geometry, Distance / midpoint.

Points $A,B,C,D$ and $E$ are located in 3-dimensional space with $AB=BC=CD=DE=EA=2$ and $\angle ABC=\angle CDE=\angle DEA=90^o$. The plane of $\triangle ABC$ is parallel to $\overline{DE}$. What is the area of $\triangle BDE$?

点 $A,B,C,D$ 和 $E$ 位于三维空间中，满足 $AB=BC=CD=DE=EA=2$ 且 $\angle ABC=\angle CDE=\angle DEA=90^o$。$\triangle ABC$ 所在平面与 $\overline{DE}$ 平行。求 $\triangle BDE$ 的面积。

(A) $\sqrt{2}$ $\sqrt{2}$

(B) $\sqrt{3}$ $\sqrt{3}$

(D) $\sqrt{5}$ $\sqrt{5}$

(E) $\sqrt{6}$ $\sqrt{6}$

Answer

Correct choice: (C)

正确答案：（C）

Solution

Let $A=(0,0,0)$, and $B=(2,0,0)$. Since $EA=2$, we could let $C=(2,0,2)$, $D=(2,2,2)$, and $E=(2,2,0)$. Now to get back to $A$ we need another vertex $F=(0,2,0)$. Now if we look at this configuration as if it was two dimensions, we would see a square missing a side if we don't draw $FA$. Now we can bend these three sides into an equilateral triangle, and the coordinates change: $A=(0,0,0)$, $B=(2,0,0)$, $C=(2,0,2)$, $D=(1,\sqrt{3},2)$, and $E=(1,\sqrt{3},0)$. Checking for all the requirements, they are all satisfied. Now we find the area of triangle $BDE$. The side lengths of this triangle are $2, 2, 2\sqrt{2}$, which is an isosceles right triangle. Thus the area of it is $\frac{2\cdot2}{2}=2\Rightarrow \mathrm{(C)}$.

设 $A=(0,0,0)$，$B=(2,0,0)$。由于 $EA=2$，可取 $C=(2,0,2)$，$D=(2,2,2)$，$E=(2,2,0)$。为了回到 $A$，再取一点 $F=(0,2,0)$。若将该构型看作二维，会看到一个缺一条边的正方形（若不画 $FA$）。现在把这三条边折成一个等边三角形，坐标变为：$A=(0,0,0)$，$B=(2,0,0)$，$C=(2,0,2)$，$D=(1,\sqrt{3},2)$，$E=(1,\sqrt{3},0)$。检验可知满足所有条件。接着求 $\triangle BDE$ 的面积。该三角形三边长为 $2, 2, 2\sqrt{2}$，是等腰直角三角形，因此面积为 $\frac{2\cdot2}{2}=2\Rightarrow \mathrm{(C)}$。

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