AMC12 2016 A

AMC12 2016 A · Q17

AMC12 2016 A · Q17. It mainly tests Triangles (properties), Transformations.

Let $ABCD$ be a square. Let $E$, $F$, $G$, and $H$ be the centers, respectively, of equilateral triangles with bases $AB$, $BC$, $CD$, and $DA$, each exterior to the square. What is the ratio of the area of square $EFGH$ to the area of square $ABCD$?

设 $ABCD$ 为正方形。分别在边 $AB$、$BC$、$CD$、$DA$ 上向正方形外侧作以这些边为底的等边三角形，其中心分别为 $E$、$F$、$G$、$H$。求正方形 $EFGH$ 的面积与正方形 $ABCD$ 的面积之比。

(A) 1 1

(B) \frac{2+\sqrt{3}}{3} \frac{2+\sqrt{3}}{3}

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

(E) \sqrt{3} \sqrt{3}

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): Without loss of generality, let the square and equilateral triangles have side length 6. Then the height of the equilateral triangles is $3\sqrt{3}$, and the distance of each of the triangle centers, $E$, $F$, $G$, and $H$, to the square $ABCD$ is $\sqrt{3}$. It follows that the diagonal of square $ABCD$ has length $6\sqrt{2}$, and the diagonal of square $EFGH$ has length equal to the side length of square $ABCD$ plus twice the distance from the center of an equilateral triangle to square $ABCD$ or $6+2\sqrt{3}$. The required ratio of the areas of the two squares is equal to the square of the ratio of the lengths of the diagonals of the two squares, or $$\left(\frac{6+2\sqrt{3}}{6\sqrt{2}}\right)^2=\left(\frac{3+\sqrt{3}}{3\sqrt{2}}\right)^2=\frac{12+6\sqrt{3}}{18}=\frac{2+\sqrt{3}}{3}.$$

答案（B）：不失一般性，设正方形和等边三角形的边长为 $6$。则等边三角形的高为 $3\sqrt{3}$，且各三角形的中心 $E$、$F$、$G$、$H$ 到正方形 $ABCD$ 的距离为 $\sqrt{3}$。因此，正方形 $ABCD$ 的对角线长为 $6\sqrt{2}$，而正方形 $EFGH$ 的对角线长度等于正方形 $ABCD$ 的边长加上等边三角形中心到正方形 $ABCD$ 的距离的两倍，即 $6+2\sqrt{3}$。所求两正方形面积之比等于它们对角线长度之比的平方，即 $$\left(\frac{6+2\sqrt{3}}{6\sqrt{2}}\right)^2=\left(\frac{3+\sqrt{3}}{3\sqrt{2}}\right)^2=\frac{12+6\sqrt{3}}{18}=\frac{2+\sqrt{3}}{3}.$$

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