Question

连接正六边形 $ABCDEF$ 各边的中点，形成一个较小的六边形。这个较小六边形所围面积是 $ABCDEF$ 面积的几分之几？

The midpoints of the sides of a regular hexagon $ABCDEF$ are joined to form a smaller hexagon. What fraction of the area of $ABCDEF$ is enclosed by the smaller hexagon?

$\frac12$

$\frac{\sqrt3}{3}$

$\frac23$

$\frac34$

$\frac{\sqrt3}{2}$

答案：D

知识点：similar polygons area ratios

难度评级：1570

小提示：

把相邻两边中点之间的距离与原六边形边长比较

Compare the distance between adjacent side midpoints with the original side length

大提示：

较小六边形与较大六边形相似，所以将边长之比平方

The smaller and larger hexagons are similar, so square their side-length ratio

解答：

两个相邻中点与它们共有的顶点组成一个三角形，其中两边均为 $\frac{s}{2}$ ，夹角为 $120^\circ$ 。其对边长度的平方为 $\begin{aligned} \frac{s^2}{4}+\frac{s^2}{4} &-2\left(\frac{s}{2}\right)^2\cos120^\circ\\ &=\frac{3s^2}{4} \end{aligned}\text{。}$ 因此较小六边形的缩放比例为 $\frac{\sqrt3}{2}$ ，面积之比为 $\frac{3}{4}$ 。正确答案是 D。

Two adjacent midpoints and their shared vertex form a triangle with two sides $\frac{s}{2}$ and included angle $120^\circ.$ Its opposite side has squared length $\begin{aligned} \frac{s^2}{4}+\frac{s^2}{4} &-2\left(\frac{s}{2}\right)^2\cos120^\circ\\ &=\frac{3s^2}{4}. \end{aligned}$ Thus the smaller hexagon has scale factor $\frac{\sqrt3}{2},$ and its area ratio is $\frac{3}{4}.$ The correct answer is D.

1996 AMC 12 第 19 题

答案：D

难度评级：1570

小提示：

大提示：

解答：

其他年份的第 19 题