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

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

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

解答：

其他年份的第 15 题