Question

正方形 $ABCD$ 的边长为 $1$ 。点 $P$ 、 $Q$ 、 $R$ 、 $S$ 分别在 $ABCD$ 的边上，使得 $APQCRS$ 是一个等边凸六边形，边长为 $s$ 。求 $s$ 。

Square $ABCD$ has side length $1.$ Points $P,$ $Q,$ $R,$ and $S$ each lie on a side of $ABCD$ such that $APQCRS$ is an equilateral convex hexagon with side length $s.$ What is $s?$

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

$\dfrac{1}{2}$

$2 - \sqrt{2}$

$1 - \dfrac{\sqrt{2}}{4}$

$\dfrac{2}{3}$

答案：C

知识点：正方形（几何）特殊直角三角形分母有理化

难度评级：1540

解答：

参考图形：

由 $AP = QC = s$ ，可知 $PB = BQ$ 。因此 $\triangle PBQ$ 是以 $PQ=s$ 为斜边的等腰直角三角形。由勾股定理， $PB = \dfrac{s}{\sqrt{2}}$ 。

在角 B 附近，三角形 PBQ 是等腰直角三角形，所以 $1 = AB = AP + PB = s + \dfrac{s}{\sqrt{2}}.$

整理可得因而 $1 = (1 + \dfrac{1}{\sqrt{2}})s$ $s = \dfrac{1}{1 + \dfrac{1}{\sqrt{2}}} = \dfrac{\sqrt{2}}{\sqrt{2} + 1}.$

将这个分式有理化，得到

$\dfrac{\sqrt{2}}{\sqrt{2} + 1} \cdot \dfrac{\sqrt{2} - 1}{\sqrt{2} - 1} = 2 - \sqrt{2}.$

所以正确答案是 C。

Consider the diagram:

Since $AP = QC = s,$ we know that $PB = BQ.$ This shows that $\triangle PBQ$ is an isosceles right triangle with hypotenuse $PQ=s.$ Using the Pythagorean theorem, we get that $PB = \dfrac{s}{\sqrt{2}}.$

We also know that $1 = AB = AP + PB = s + \dfrac{s}{\sqrt{2}}.$

This equation simplifies to $1 = (1 + \dfrac{1}{\sqrt{2}})s$ Which implies that $s = \dfrac{1}{1 + \dfrac{1}{\sqrt{2}}} = \dfrac{\sqrt{2}}{\sqrt{2} + 1}.$

We can rationalize this fraction to get

$\dfrac{\sqrt{2}}{\sqrt{2} + 1} \cdot \dfrac{\sqrt{2} - 1}{\sqrt{2} - 1} = 2 - \sqrt{2}.$

Thus, C is the correct answer.

2022 AMC 10A 第 5 题

答案：C

难度评级：1540

解答：

其他年份的第 5 题