2021 AMC 12B Fall 第 15 题

Question

三张相同的边长为

6

的正方形纸片叠在一起。中间的纸片绕其中心顺时针旋转

30^\circ

，最上面的纸片绕其中心顺时针旋转

60^\circ

，得到下图所示的

24

边形。该多边形面积可表示为

a - b\sqrt{c}

，其中

a

、

b

、

c

为正整数，且

c

不被任何质数的平方整除。求

a + b + c

。

Three identical square sheets of paper each with side length

6

are stacked on top of each other. The middle sheet is rotated clockwise

30^\circ

about its center and the top sheet is rotated clockwise

60^\circ

about its center, resulting in the

24

-sided polygon shown in the figure below. The area of this polygon can be expressed in the form

a - b\sqrt{c},

where

a,

b,

and

c

are positive integers, and

c

is not divisible by the square of any prime. What is

a + b + c?

75

93

96

129

147

答案：E

知识点：面积分割三角学对称性

难度评级：2100

小提示：

由对称性，多边形的顶点每隔

15^\circ

交替为外侧正方形顶点和内部边交点。

By symmetry the polygon’s vertices alternate between outer square-corners and inner edge-crossings, every

15^\circ

大提示：

外侧顶点到中心的距离为

3\sqrt2

，内部顶点到中心的距离为

2\sqrt3

。

The outer vertices are at distance

3\sqrt2

from the center and the inner ones at distance

2\sqrt3

解答：

因为三个正方形分别旋转

0^\circ

，

30^\circ

，

60^\circ

，图形有

12

重对称。它的

24

个顶点每隔

15^\circ

交替出现：外侧顶点到中心的距离为

3\sqrt2

，内侧交点到中心的距离为

2\sqrt3

。

连接中心与全部

24

个顶点，把多边形分成

24

个三角形，每个三角形的两边为

3\sqrt2

和

2\sqrt3

，夹角为

15^\circ

。总面积是

\begin{aligned} &24 \cdot \tfrac12 (3\sqrt2)(2\sqrt3)\sin 15^\circ \\ &= 72\sqrt6 \cdot \dfrac{\sqrt6 - \sqrt2}{4} \end{aligned}\text{。}

这化简为

18\sqrt6(\sqrt6 - \sqrt2) = 108 - 36\sqrt3

，所以

a + b + c = 108 + 36 + 3 = 147

。

所以正确答案是 E。

Because the three squares are rotated by

0^\circ,

30^\circ,

and

60^\circ,

the figure has

12

-fold symmetry. Its

24

vertices alternate every

15^\circ

: outer vertices are the square corners at distance

3\sqrt2

from the center, and inner vertices are edge crossings at distance

2\sqrt3.

Connecting the center to all

24

vertices splits the polygon into

24

triangles, each with sides

3\sqrt2

and

2\sqrt3

and included angle

15^\circ.

The total area is

\begin{aligned} &24 \cdot \tfrac12 (3\sqrt2)(2\sqrt3)\sin 15^\circ \\ &= 72\sqrt6 \cdot \dfrac{\sqrt6 - \sqrt2}{4}. \end{aligned}

This simplifies to

18\sqrt6(\sqrt6 - \sqrt2) = 108 - 36\sqrt3,

so

a + b + c = 108 + 36 + 3 = 147.

Thus, the correct answer is E.

2021 AMC 12B Fall 第 15 题

答案：E

难度评级：2100

小提示：

大提示：

解答：

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