2008 AMC 12B 第 15 题

Question

在一个单位正方形的每条边上，向外作一个边长为

1

的等边三角形。在每个等边三角形的新边上，再作一个边长为

1

的等边三角形。正方形和这

12

个三角形的内部没有公共点。令

R

为正方形和所有三角形的并集，令

S

为包含

R

的最小凸多边形。位于

S

内部但在

R

外部的区域面积是多少？

On each side of a unit square, an equilateral triangle of side length

1

is constructed. On each new side of each equilateral triangle, another equilateral triangle of side length

1

is constructed. The interiors of the square and the

12

triangles have no points in common. Let

R

be the region formed by the union of the square and all the triangles, and let

S

be the smallest convex polygon that contains

R.

What is the area of the region that is inside

S

but outside

R?

\dfrac{1}{4}

\dfrac{\sqrt{2}}{4}

1

\sqrt{3}

2\sqrt{3}

答案：C

知识点：三角形面积导角三角学

难度评级：1660

小提示：

S

与

R

之间的空隙是四个小三角形，正方形的每个角各一个。

The gap between

S

and

R

is four small triangles, one at each corner of the square

大提示：

每个空隙三角形有两条长度为

1

的边，夹角为

360^\circ - 90^\circ - 4 \cdot 60^\circ

；使用

\tfrac12 ab \sin C

。

Each gap triangle has two sides of length

1

meeting at angle

360^\circ - 90^\circ - 4 \cdot 60^\circ;

use

\tfrac12 ab \sin C

解答：

凸包

S

与

R

的差只出现在正方形的四个角附近，每处形成一个小三角形空隙。每个空隙三角形有两条长度为

1

的边，也就是相邻三角形的外边。

这两条边之间的夹角为

360^\circ - 90^\circ - 4 \cdot 60^\circ = 30^\circ

，所以每个空隙面积为

\tfrac12 \cdot 1 \cdot 1 \cdot \sin 30^\circ = \tfrac14\text{。}

总面积为

4 \cdot \tfrac14 = 1

。

因此，正确答案是 C。

The convex hull

S

differs from

R

only near the four corners of the square, where a small triangular gap forms. Each gap triangle has two sides of length

1

(outer edges of adjacent triangles).

The angle between those two sides is

360^\circ - 90^\circ - 4 \cdot 60^\circ = 30^\circ,

so each gap has area

\tfrac12 \cdot 1 \cdot 1 \cdot \sin 30^\circ = \tfrac14.

The total area is

4 \cdot \tfrac14 = 1.

Thus, the correct answer is C.

2008 AMC 12B 第 15 题

答案：C

难度评级：1660

小提示：

大提示：

解答：

其他年份的第 15 题