AMC10 2023 A

AMC10 2023 A · Q24

AMC10 2023 A · Q24. It mainly tests Area & perimeter, Polygons.

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?

六个边长为 $1$ 单位的正六边形木块排列在一个正六边形框架内。每个木块位于框架内侧边上，并与另外两个木块对齐，如图所示。从框架的任意角到最近的木块顶点的距离为 $\frac{3}{7}$ 单位。求框架内未被木块占据的区域面积。

(A) \frac{13 \sqrt{3}}{3} \frac{13 \sqrt{3}}{3}

(B) \frac{216 \sqrt{3}}{49} \frac{216 \sqrt{3}}{49}

(D) \frac{14 \sqrt{3}}{3} \frac{14 \sqrt{3}}{3}

(E) \frac{243 \sqrt{3}}{49} \frac{243 \sqrt{3}}{49}

Answer

Correct choice: (C)

正确答案：（C）

Solution

Examining the red isosceles trapezoid with $1$ and $\dfrac{3}{7}$ as two bases, we know that the side lengths are $\dfrac{4}{7}$ from $30-60-90$ triangle. We can conclude that the big hexagon has side length 3. Thus the target area is: area of the big hexagon - 6 * area of the small hexagon. $\dfrac{3\sqrt{3}}{2}(3^2-6\cdot1^2) = \dfrac{3\sqrt{3}}{2}(3) = \boxed{\textbf{(C)}~\frac{9 \sqrt{3}}{2}}$ We can extend the line of the parallelogram to the end until it touches the next hexagon and it will make a small equilateral triangle and a longer parallelogram. We can prove that one side of the tiny equilateral triangle is 4/7 by playing around with angles and the parallelogram because it is parallel, we can then use the whole side of the hexagon which is one and subtract 3/7 which is one side of the equilateral triangle which is 4/7. That means the whole side of the big hexagon length is 3 and we can continue with solution 1.

考察红色等腰梯形，其两底分别为 $1$ 和 $\dfrac{3}{7}$，由 $30-60-90$ 三角形知其腿长为 $\dfrac{4}{7}$。由此可知大六边形边长为 $3$。目标面积为：大六边形面积减去 $6$ 个小六边形面积。 $\dfrac{3\sqrt{3}}{2}(3^2-6\cdot1^2) = \dfrac{3\sqrt{3}}{2}(3) = \boxed{\textbf{(C)}~\frac{9 \sqrt{3}}{2}}$ 还可以将平行四边形线延长至接触下一个六边形，形成一个小正三角形和较长的平行四边形。通过角度和平行四边形性质，可证小正三角形边长为 $4/7$。则大六边形全边长为 $1$，减去 $\frac{3}{7}$（小三角形一侧，为 $4/7$？），大六边形边长为 $3$，继续解法一。

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