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

小提示：

未覆盖区域等于框架面积减去六个单位正六边形面积；边长为 $t$ 的正六边形面积为 $\tfrac{3\sqrt3}{2}t^2$

The uncovered region is the frame’s area minus the six unit hexagons; a regular hexagon of side $t$ has area $\tfrac{3\sqrt3}{2}t^2$

大提示：

把积木的边延长到框架：框架的一条边被分成长度 $\tfrac37, 1, 1$ 和 $1-\tfrac37$ ；然后减去六个积木的面积

Extend the block edges to the frame: one frame side splits into lengths $\tfrac37, 1, 1,$ and $1-\tfrac37;$ then subtract the six block areas

解答：

设 $d=\tfrac37$ 。把与框架某条固定边相接的那些积木的斜边延长。由于所有相关的角都是 $60^\circ$ ，这些延长线在一端围成一个边长为 $1$ 的正三角形，在另一端围成一个边长为 $1-d$ 的正三角形。于是这条框架边被分成长度依次为 $d,1,1$ 和 $1-d$ 的四段，所以它的长度是 $d+1+1+(1-d)=3$ 。边长为 $t$ 的正六边形面积为 $\tfrac{3\sqrt3}{2}t^2$ 。因此未被占据的面积等于边长为 $3$ 的框架面积减去六个单位积木的面积： $\tfrac{3\sqrt3}{2}\cdot 3^2$ $- 6\cdot\tfrac{3\sqrt3}{2}$ $= \tfrac{27\sqrt3}{2} - 9\sqrt3$ $= \tfrac{9\sqrt3}{2}$ 。因此，答案是 C。

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

小提示：

大提示：

解答：

其他年份的第 24 题