Question

设 $ABCDE$ 是周长为 $1$ 的等角凸五边形。延长五边形各边所得直线的两两交点形成一个五角星多边形。设这个五角星的周长为 $s$ 。 $s$ 的最大可能值与最小可能值之差是多少？

Let $ABCDE$ be an equiangular convex pentagon of perimeter $1.$ The pairwise intersections of the lines that extend the sides of the pentagon determine a five-pointed star polygon. Let $s$ be the perimeter of this star. What is the difference between the maximum and the minimum possible values of $s?$

$0$

$\dfrac{1}{2}$

$\dfrac{\sqrt5 - 1}{2}$

$\dfrac{\sqrt5 + 1}{2}$

$\sqrt5$

答案：A

知识点：等角多边形等腰三角形不变量

难度评级：1890

解答：

等角五边形的内角全为 $108^\circ$ ，所以星形的每个尖角都是底角为 $72^\circ$ 、顶角为 $36^\circ$ 的等腰三角形。由于底角相等，每个尖角贡献的两条边都是其所对应五边形边长的同一固定倍数 $c$ 。把五个尖角相加，星形周长为 $2c\cdot(\text{pentagon perimeter}) = 2c$ ，与各边的具体长度无关。因此 $s$ 是常数，其最大值与最小值之差为 $0$ 。所以正确答案是 A。

An equiangular pentagon has all interior angles $108^\circ,$ so each point of the star is an isosceles triangle with base angles $72^\circ$ and apex $36^\circ.$ By the equal base angles, each point contributes two sides that are the same fixed multiple $c$ of the pentagon side it rests on. Summing over the five points, the star perimeter equals $2c\cdot(\text{pentagon perimeter}) = 2c,$ independent of the individual side lengths. So $s$ is constant, and the difference between its maximum and minimum values is $0.$ Thus, the correct answer is A.

2013 AMC 12B 第 16 题

答案：A

难度评级：1890

解答：

其他年份的第 16 题