2018 AMC 12B 第 20 题

Question

设

ABCDEF

是边长为

1

的正六边形。记

X

、

Y

和

Z

分别为边

AB

、

CD

和

EF

的中点。内部区域为

\triangle ACE

与

\triangle XYZ

的内部交集的凸六边形面积是多少？

Let

ABCDEF

be a regular hexagon with side length

1.

Denote by

X,

Y,

and

Z

the midpoints of sides

AB,

CD,

and

EF,

respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of

\triangle ACE

and

\triangle XYZ?

\dfrac{3}{8}\sqrt{3}

\dfrac{7}{16}\sqrt{3}

\dfrac{15}{32}\sqrt{3}

\dfrac{1}{2}\sqrt{3}

\dfrac{9}{16}\sqrt{3}

答案：C

知识点：正多边形面积比等边三角形

难度评级：2270

小提示：

把这个正六边形放在单位圆上，写出

A,C,E

和三个中点的坐标。

Put the regular hexagon on the unit circle and write coordinates for

A,C,E

and the three midpoints

大提示：

求这两个等边三角形各边所在直线的交点，得到六个顶点，再使用鞋带公式。

Intersect the side lines of the two equilateral triangles to find the six vertices, then use shoelace

解答：

把正六边形放在单位圆上，取

A=(1,0)

、

C=(-\tfrac12,\tfrac{\sqrt3}{2})

和

E=(-\tfrac12,-\tfrac{\sqrt3}{2})

。题中指定的三个中点为

X=(\tfrac34,\tfrac{\sqrt3}{4})

、

Y=(-\tfrac34,\tfrac{\sqrt3}{4})

和

Z=(0,-\tfrac{\sqrt3}{2})

。

求

\triangle ACE

与

\triangle XYZ

各边所在直线的交点，就得到公共内部的六个顶点，按循环顺序为

\begin{gathered} (-\tfrac12,0),\quad (-\tfrac18,-\tfrac{3\sqrt3}{8}),\\ (\tfrac14,-\tfrac{\sqrt3}{4}),\quad (\tfrac58,\tfrac{\sqrt3}{8}),\\ (\tfrac14,\tfrac{\sqrt3}{4}),\quad (-\tfrac12,\tfrac{\sqrt3}{4}) \end{gathered}\text{。}

对这些顶点使用鞋带公式，得到面积为

\dfrac{15\sqrt3}{32}

。

所以正确答案是 C。

Place the regular hexagon on the unit circle with

A=(1,0),

C=(-\tfrac12,\tfrac{\sqrt3}{2}),

and

E=(-\tfrac12,-\tfrac{\sqrt3}{2}).

The three specified midpoints are

X=(\tfrac34,\tfrac{\sqrt3}{4}),

Y=(-\tfrac34,\tfrac{\sqrt3}{4}),

and

Z=(0,-\tfrac{\sqrt3}{2}).

Intersecting the side lines of

\triangle ACE

and

\triangle XYZ

gives the six vertices of their common interior, in cyclic order:

\begin{gathered} (-\tfrac12,0),\quad (-\tfrac18,-\tfrac{3\sqrt3}{8}),\\ (\tfrac14,-\tfrac{\sqrt3}{4}),\quad (\tfrac58,\tfrac{\sqrt3}{8}),\\ (\tfrac14,\tfrac{\sqrt3}{4}),\quad (-\tfrac12,\tfrac{\sqrt3}{4}). \end{gathered}

The shoelace formula applied to these vertices gives area

\dfrac{15\sqrt3}{32}.

Thus, the correct answer is C.

2018 AMC 12B 第 20 题

答案：C

难度评级：2270

小提示：

大提示：

解答：

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