Question

六个边长为 $1$ 单位的正六边形积木排列在一个正六边形框架内。每个积木都沿着框架的一条内边放置，并与另外两个积木对齐，如下图所示。从框架任意顶点到最近的积木顶点的距离都是 $\frac{3}{7}$ 单位。框架内未被积木占据的区域面积是多少？

Six regular hexagonal blocks of side length $1$ unit are arranged inside a regular hexagonal frame. Each block lies along an inside edge of the frame and is aligned with two other blocks, as shown in the figure below. The distance from any corner of the frame to the nearest vertex of a block is $\frac{3}{7}$ unit. What is the area of the region inside the frame not occupied by the blocks?

$\dfrac{13\sqrt{3}}{3}$

$\dfrac{216\sqrt{3}}{49}$

$\dfrac{9\sqrt{3}}{2}$

$\dfrac{14\sqrt{3}}{3}$

$\dfrac{243\sqrt{3}}{49}$

答案：C

知识点：正多边形面积面积分割

难度评级：2520

解答：

未覆盖区域等于框架面积减去六个单位积木面积。边长为 $t$ 的正六边形面积为 $\tfrac{3\sqrt3}{2}t^2$ ，所以每个单位积木面积为 $d=\tfrac37$ 。每个框架顶点到最近积木顶点的距离都是 $60^\circ$ ，这个间距条件确定框架边长为 $3$ 。因此未覆盖面积为 $\tfrac{3\sqrt3}{2}\cdot 3^2$ $- 6\cdot\tfrac{3\sqrt3}{2}$ $= \tfrac{27\sqrt3}{2} - 9\sqrt3$ $1$ 。因此，答案是 C。 $1-d$ $d,1,1$ ， $1-d$ ， $d+1+1+(1-d)=3$ 。 $= \tfrac{9\sqrt3}{2}$ 。

Let $d=\tfrac37.$ Extend the slanted edges of the blocks that meet a fixed side of the frame. Because all the relevant angles are $60^\circ,$ the extensions form an equilateral triangle of side $1$ at one end and an equilateral triangle of side $1-d$ at the other. Thus that frame side is partitioned into lengths $d,1,1,$ and $1-d,$ so its length is $d+1+1+(1-d)=3.$ A regular hexagon of side $t$ has area $\tfrac{3\sqrt3}{2}t^2.$ Therefore the uncovered area is the area of the side- $3$ frame minus the areas of the six unit blocks: $\tfrac{3\sqrt3}{2}\cdot 3^2$ $- 6\cdot\tfrac{3\sqrt3}{2}$ $= \tfrac{27\sqrt3}{2} - 9\sqrt3$ $= \tfrac{9\sqrt3}{2}.$ Therefore, the answer is C.

2023 AMC 10A 第 24 题

答案：C

难度评级：2520

解答：

其他年份的第 24 题