Question

四个正六边形围绕一个边长为 $1$ 的正方形，每个六边形都与该正方形共用一条边，如下图所示。所得外侧 $12$ 边非凸多边形的面积可写成 $m\sqrt n + p$ ，其中 $m$ 、 $n$ 和 $p$ 都是整数，且 $n$ 不被任何质数的平方整除。求 $m + n + p$ 。

Four regular hexagons surround a square with a side length $1,$ each one sharing an edge with the square, as shown in the figure below. The area of the resulting $12$ -sided outer nonconvex polygon can be written as $m\sqrt n + p,$ where $m,$ $n,$ and $p$ are integers and $n$ is not divisible by the square of any prime. What is $m + n + p?$

$-12$

$-4$

$4$

$24$

$32$

答案：B

知识点：正多边形坐标几何鞋带公式

难度评级：2520

解答：

将正方形中心置于原点，顶点为 $\left(\pm\tfrac12, \pm\tfrac12\right)$ 。每个六边形与正方形共用一条边，并向正方形的对侧延伸。例如底边上的六边形，其远端上边从 $\left(-\tfrac12, \sqrt3 - \tfrac12\right)$ 到 $\left(\tfrac12, \sqrt3 - \tfrac12\right)$ 。

外边界是一个 $12$ 边形，其平边到中心的距离为 $\sqrt3 - \tfrac12$ ，凸顶点包括 $\left(\sqrt3 - \tfrac12, \tfrac12\right)$ ，而相邻六边形的斜边在 $\left(\tfrac52 - \sqrt3, \tfrac52 - \sqrt3\right)$ 及其对称点相交，形成四个凹入的顶点。

对这 $12$ 个顶点使用鞋带公式，面积为 $16\sqrt3 - 23$ ，所以 $m = 16$ ， $n = 3$ ， $p = -23$ ，从而 $m + n + p = -4$ 。

所以正确答案是 B。

Center the square at the origin with vertices $\left(\pm\tfrac12, \pm\tfrac12\right).$ Each hexagon shares one edge with the square and extends across to the opposite side; the hexagon on the bottom edge, for instance, has its far (top) edge from $\left(-\tfrac12, \sqrt3 - \tfrac12\right)$ to $\left(\tfrac12, \sqrt3 - \tfrac12\right).$

The outer boundary is a $12$ -gon with flat edges at distance $\sqrt3 - \tfrac12$ from the center, convex vertices such as $\left(\sqrt3 - \tfrac12, \tfrac12\right),$ and four reflex notches where adjacent hexagons' slanted edges meet, at $\left(\tfrac52 - \sqrt3, \tfrac52 - \sqrt3\right)$ and its symmetric images.

Applying the shoelace formula to these $12$ vertices gives area $16\sqrt3 - 23,$ so $m = 16,$ $n = 3,$ $p = -23,$ and $m + n + p = -4.$

Thus, the correct answer is B.

2022 AMC 12B 第 25 题

答案：B

难度评级：2520

解答：

其他年份的第 25 题