AMC12 2022 B

AMC12 2022 B · Q25

AMC12 2022 B · Q25. It mainly tests Area & perimeter, Polygons.

Four regular hexagons surround a square with 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 $m+n+p$?

四个正六边形围绕着一个边长为 $1$ 的正方形，每个六边形与正方形共享一条边，如下图所示。所得 $12$ 边外部非凸多边形的面积可写成 $m \sqrt{n} + p$，其中 $m$、$n$、$p$ 为整数，且 $n$ 任何质数的平方都不整除 $n$。 $m+n+p$ 是多少？

(A) -12 -12

(B) -4 -4

(D) 24 24

(E) 32 32

Answer

Correct choice: (B)

正确答案：（B）

Solution

Refer to the diagram above. Let the origin be at the center of the square, $A$ be the intersection of the top and right hexagons, $B$ be the intersection of the top and left hexagons, and $M$ and $N$ be the top points in the diagram. By symmetry, $A$ lies on the line $y = x$. The equation of line $AN$ is $y = -x\sqrt{3} + \frac{3}{2}\sqrt{3} - \frac{1}{2}$ (due to it being one of the sides of the top hexagon). Thus, we can solve for the coordinates of $A$ by finding the intersection of the two lines: \[x = -x\sqrt{3} + \frac{3\sqrt{3} - 1}{2}\] \[x(\sqrt{3} + 1) = \frac{3\sqrt{3} - 1}{2}\] \[x = \frac{3\sqrt{3}-1}{2} \cdot \frac{1}{\sqrt{3} + 1}\] \[= \frac{3\sqrt{3}-1}{2(\sqrt{3} + 1)} \cdot \frac{\sqrt{3} - 1}{\sqrt{3} - 1}\] \[= \frac{10 - 4\sqrt{3}}{4}\] \[= \frac{5}{2} - \sqrt{3}\] \[\therefore A = \left(\frac{5}{2} - \sqrt{3}, \frac{5}{2} - \sqrt{3}\right).\] This means that we can find the length $AB$, which is equal to $2(\frac{5}{2} - \sqrt{3}) = 5 - 2\sqrt{3}$. We will next find the area of trapezoid $ABMN$. The lengths of the bases are $1$ and $5 - 2\sqrt{3}$, and the height is equal to the $y$-coordinate of $M$ minus the $y$-coordinate of $A$. The height of the hexagon is $\sqrt{3}$ and the bottom of the hexagon lies on the line $y = \frac{1}{2}$. Thus, the $y$-coordinate of $M$ is $\sqrt{3} - \frac{1}{2}$, and the height is $2\sqrt{3} - 3$. We can now find the area of the trapezoid: \[[ABMN] = (2\sqrt{3} - 3)\left(\frac{1 + 5 - 2\sqrt{3}}{2}\right)\] \[= (2\sqrt{3} - 3)(3 - \sqrt{3})\] \[= 6\sqrt{3} + 3\sqrt{3} - 9 - 6\] \[= 9\sqrt{3} - 15.\] The total area of the figure is the area of a square with side length $AB$ plus four times the area of this trapezoid: \[\textrm{Area} = (5 - 2\sqrt{3})^2 + 4(9\sqrt{3} - 15)\] \[= 37 - 20\sqrt{3} + 36\sqrt{3} - 60\] \[= 16\sqrt{3} - 23.\] Our answer is $16 + 3 - 23 = \boxed{\textbf{(B) }-4}$.

参考上图。设原点为正方形的中心，$A$ 为顶部和右侧六边形的交点，$B$ 为顶部和左侧六边形的交点，$M$ 和 $N$ 为图中的顶部点。由对称性，$A$ 位于直线 $y = x$ 上。直线 $AN$ 的方程为 $y = -x\sqrt{3} + \frac{3}{2}\sqrt{3} - \frac{1}{2}$（因为它是顶部六边形的一条边）。因此，通过求两条直线的交点可以得到 $A$ 的坐标： \[x = -x\sqrt{3} + \frac{3\sqrt{3} - 1}{2}\] \[x(\sqrt{3} + 1) = \frac{3\sqrt{3} - 1}{2}\] \[x = \frac{3\sqrt{3}-1}{2} \cdot \frac{1}{\sqrt{3} + 1}\] \[= \frac{3\sqrt{3}-1}{2(\sqrt{3} + 1)} \cdot \frac{\sqrt{3} - 1}{\sqrt{3} - 1}\] \[= \frac{10 - 4\sqrt{3}}{4}\] \[= \frac{5}{2} - \sqrt{3}\] \[= \left(\frac{5}{2} - \sqrt{3}, \frac{5}{2} - \sqrt{3}\right).\] 这意味着我们可以求出长度 $AB$，等于 $2(\frac{5}{2} - \sqrt{3}) = 5 - 2\sqrt{3}$。接下来求梯形 $ABMN$ 的面积。底边长度分别为 $1$ 和 $5 - 2\sqrt{3}$，高度等于 $M$ 的 $y$ 坐标减去 $A$ 的 $y$ 坐标。六边形的高度为 $\sqrt{3}$，六边形的底部位于直线 $y = \frac{1}{2}$ 上。因此，$M$ 的 $y$ 坐标为 $\sqrt{3} - \frac{1}{2}$，高度为 $2\sqrt{3} - 3$。现在可以求梯形的面积： \[[ABMN] = (2\sqrt{3} - 3)\left(\frac{1 + 5 - 2\sqrt{3}}{2}\right)\] \[= (2\sqrt{3} - 3)(3 - \sqrt{3})\] \[= 6\sqrt{3} + 3\sqrt{3} - 9 - 6\] \[= 9\sqrt{3} - 15.\] 图形的总面积为边长为 $AB$ 的正方形的面积加上该梯形面积的四倍： \[\textrm{Area} = (5 - 2\sqrt{3})^2 + 4(9\sqrt{3} - 15)\] \[= 37 - 20\sqrt{3} + 36\sqrt{3} - 60\] \[= 16\sqrt{3} - 23.\] 答案为 $16 + 3 - 23 = \boxed{\textbf{(B) }-4}$。

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