AMC10 2006 A

AMC10 2006 A · Q17

AMC10 2006 A · Q17. It mainly tests Area & perimeter, Coordinate geometry.

In rectangle ADEH, points B and C trisect AD, and points G and F trisect HE. In addition, AH = AC = 2. What is the area of quadrilateral WXYZ shown in the figure?

在矩形 $ADEH$ 中，点 $B$ 和 $C$ 将 $AD$ 三等分，点 $G$ 和 $F$ 将 $HE$ 三等分。此外，$AH = AC = 2$。如图所示四边形 $WXYZ$ 的面积是多少？

(A) $\frac{1}{2}$ $\frac{1}{2}$

(B) $\frac{\sqrt{2}}{2}$ $\frac{\sqrt{2}}{2}$

(D) $\frac{2\sqrt{2}}{3}$ $\frac{2\sqrt{2}}{3}$

(E) $\frac{2\sqrt{3}}{3}$ $\frac{2\sqrt{3}}{3}$

Answer

Correct choice: (A)

正确答案：（A）

Solution

(A) First note that since points $B$ and $C$ trisect $\overline{AD}$, and points $G$ and $F$ trisect $\overline{HE}$, we have $HG = GF = FE = AB = BC = CD = 1$. Also, $HG$ is parallel to $CD$ and $HG = CD$, so $CDGH$ is a parallelogram. Similarly, $AB$ is parallel to $FE$ and $AB = FE$, so $ABEF$ is a parallelogram. As a consequence, $WXYZ$ is a parallelogram, and since $HG = CD = AB = FE$, it is a rhombus. Since $AH = AC = 2$, the rectangle $ACFH$ is a square of side length $2$. Its diagonals $\overline{AF}$ and $\overline{CH}$ have length $2\sqrt{2}$ and form a right angle at $X$. As a consequence, $WXYZ$ is a square. In isosceles $\triangle HXF$ we have $HX = XF = \sqrt{2}$. In addition, $HG = \frac{1}{2}HF$. So $XW = \frac{1}{2}XF = \frac{1}{2}\sqrt{2}$, and the square $WXYZ$ has area $XW^2 = \frac{1}{2}$.

（A）首先注意到，由于点 $B$ 和 $C$ 将 $\overline{AD}$ 三等分，而点 $G$ 和 $F$ 将 $\overline{HE}$ 三等分，所以有 $HG = GF = FE = AB = BC = CD = 1$。另外，$HG$ 平行于 $CD$ 且 $HG = CD$，因此 $CDGH$ 是一个平行四边形。类似地，$AB$ 平行于 $FE$ 且 $AB = FE$，所以 $ABEF$ 是一个平行四边形。因此，$WXYZ$ 是一个平行四边形，并且由于 $HG = CD = AB = FE$，它是一个菱形。由于 $AH = AC = 2$，矩形 $ACFH$ 是边长为 $2$ 的正方形。其对角线 $\overline{AF}$ 和 $\overline{CH}$ 的长度为 $2\sqrt{2}$，并在点 $X$ 处成直角。因此，$WXYZ$ 是一个正方形。在等腰三角形 $\triangle HXF$ 中，有 $HX = XF = \sqrt{2}$。另外，$HG = \frac{1}{2}HF$。所以 $XW = \frac{1}{2}XF = \frac{1}{2}\sqrt{2}$，并且正方形 $WXYZ$ 的面积为 $XW^2 = \frac{1}{2}$。

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