Question

一个碗由四个边长为 $1$ 的正六边形连接到一个边长为 $1$ 的正方形上形成。相邻六边形的边重合，如图所示。连接这四个六边形顶部边缘上的八个顶点得到一个八边形。求这个八边形的面积。

A bowl is formed by attaching four regular hexagons of side $1$ to a square of side $1.$ The edges of the adjacent hexagons coincide, as shown in the figure. What is the area of the octagon obtained by joining the top eight vertices of the four hexagons, situated on the rim of the bowl?

$6$

$7$

$5 + 2 \sqrt{2}$

$8$

$9$

答案：B

知识点：立体几何正多边形面积分割

难度评级：2390

解答：

从正上方观察碗口。把底部正方形放在 $(\pm\tfrac12,\pm\tfrac12,0).$ 正六边形中，与所接边相邻的一条边，其平行于该边和垂直于该边的分量分别为 $\tfrac12$ 与 $\tfrac{\sqrt3}{2}$ 。

在底部正方形的一个顶点处，相邻两个六边形的对应边重合。比较它们的水平分量可知，在六边形平面内，垂直于所接边的单位向量的水平分量为 $1/\sqrt3.$ 正六边形的对边与所接边相距 $\sqrt3$ ，所以水平向外位移为 $1.$

因此，碗口的四条单位长度边都位于底部正方形对应边之外一个单位处。其俯视图是一个 $3$ 乘 $3$ 的正方形，去掉四个直角边为 $1$ 的等腰直角三角形：

面积为 $3^2-4\left(\frac12\cdot1^2\right)=7.$

所以正确答案是 B。

View the rim from directly above. Place the bottom square at $(\pm\tfrac12,\pm\tfrac12,0).$ In a regular hexagon, an edge adjacent to the attached side has components $\tfrac12$ parallel and $\tfrac{\sqrt3}{2}$ perpendicular to that side.

At a corner of the bottom square, the corresponding edges of two adjacent hexagons coincide. Comparing their horizontal components shows that the horizontal component of a unit vector perpendicular to an attached side within its hexagon is $1/\sqrt3.$ The opposite side of a regular hexagon is $\sqrt3$ units from the attached side, so its horizontal outward displacement is $1.$

Consequently, the four unit-length sides of the rim lie one unit beyond the four sides of the bottom square. Its top view is therefore a $3$ -by- $3$ square with four isosceles right corner triangles of leg $1$ removed:

Its area is $3^2-4\left(\frac12\cdot1^2\right)=7.$

Thus, B is the correct answer.

2022 AMC 10A 第 21 题

答案：B

难度评级：2390

解答：

其他年份的第 21 题