2022 AMC 10A 第 5 题

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

小提示：

利用正方形角上的

45

–

45

–

90

三角形

Use the

45

–

45

–

90

triangle at a corner of the square

大提示：

用一个六边形边长加一个角上的小直角三角形边长表示正方形边长

Write the side of the square as one hexagon side plus one corner leg

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}}

。

我们还知道

1 = AB = AP + PB = s + \dfrac{s}{\sqrt{2}}\text{。}

整理可得

1 = (1 + \dfrac{1}{\sqrt{2}})s

因而

s = \dfrac{1}{1 + \dfrac{1}{\sqrt{2}}} = \dfrac{\sqrt{2}}{\sqrt{2} + 1}\text{。}

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}\text{。}

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

小提示：

大提示：

解答：

其他年份的第 5 题