1996 AMC 12 第 19 题

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}

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

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

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.

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