Question

一个面积为 $1 + \sqrt{5}$ 的正五边形印在纸上并剪下。将五边形的五个顶点都折到五边形的中心，形成一个较小的五边形。新五边形的面积是多少？

A regular pentagon with area $1 + \sqrt{5}$ is printed on paper and cut out. All five vertices are folded to the center of the pentagon, creating a smaller pentagon. What is the area of the new pentagon?

$4 - \sqrt{5}$

$\sqrt{5} - 1$

$8 - 3\sqrt{5}$

$\dfrac{\sqrt{5} + 1}{2}$

$\dfrac{2 + \sqrt{5}}{3}$

答案：B

知识点：正多边形折纸相似面积比

难度评级：2600

解答：

设原正五边形外接圆半径为 $R$ 。将一个顶点折到中心时，折痕是中心到该顶点线段的垂直平分线，距离中心 $\tfrac{R}{2}$ 。五条折痕围成新的正五边形，其内切半径为 $\tfrac{R}{2}$ 。原正五边形内切半径为 $R\cos 36^\circ$ ，所以新旧正五边形相似比为 $\tfrac{R/2}{R\cos 36^\circ}$ ，面积缩放因子为 $\left(\tfrac{1}{2\cos 36^\circ}\right)^2$ 。代入 $\cos 36^\circ = \tfrac{1+\sqrt5}{4}$ ： $\tfrac{2}{3+\sqrt5} = \tfrac{3-\sqrt5}{2}$ 。因此新面积为 $(\sqrt5+1)\cdot\tfrac{3-\sqrt5}{2} = \sqrt{5} - 1$ 。所以正确答案是 B。

Let the original pentagon have circumradius $R.$ Folding a vertex to the center creases along the perpendicular bisector of the center-to-vertex segment, a line at distance $\tfrac{R}{2}$ from the center. Those five creases bound the new regular pentagon, whose apothem is $\tfrac{R}{2}$ (the original apothem was $R\cos 36^\circ$ ). So the new pentagon is similar with ratio $\tfrac{R/2}{R\cos 36^\circ},$ and its area is the old area times $\left(\tfrac{1}{2\cos 36^\circ}\right)^2.$ Plug in $\cos 36^\circ = \tfrac{1+\sqrt5}{4}:$ the factor becomes $\tfrac{2}{3+\sqrt5} = \tfrac{3-\sqrt5}{2}.$ So the new area is $(\sqrt5+1)\cdot\tfrac{3-\sqrt5}{2} = \sqrt{5} - 1.$ Thus, B is the correct answer.

2023 AMC 10B 第 25 题

答案：B

难度评级：2600

解答：

其他年份的第 25 题