2014 AMC 12B 第 21 题

先试着解答 2014 AMC 12B 第 21 题，然后核对你的答案与精心整理的解答，解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2014 AMC 12B 解答，或核对答案。

所有题目均经美国数学协会（MAA）官方合法授权使用。

21.

图中，

ABCD

是边长为

1

的正方形。矩形

JKHG

和

EBCF

全等。

BE

是多少？

In the figure,

ABCD

is a square of side length

1.

The rectangles

JKHG

and

EBCF

are congruent. What is

BE?

\dfrac12(\sqrt{6} - 2)

\dfrac14

2 - \sqrt{3}

\dfrac{\sqrt{3}}{6}

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

答案：C

知识点：直角三角形三角学导角

难度评级：2350

小提示：

设

\theta = \angle DHG

；在直角三角形

GDH

中，

BE = \dfrac{1 - \cos\theta}{\sin\theta}

。

Let

\theta = \angle DHG;

in right triangle

GDH,

BE = \dfrac{1 - \cos\theta}{\sin\theta}

大提示：

沿上边，

CD = CF + FH + HD = 1

可化简为

2\sin\theta = 1

。

Along the top side,

CD = CF + FH + HD = 1

simplifies to

2\sin\theta = 1

解答：

设

x = BE = GH = CF

，并设

\theta = \angle DHG = \angle AGJ

= \angle FKH

，其中

AD = GJ = HK = 1

。在直角三角形

GDH

中，

x \sin\theta = DG = 1 - \cos\theta

，所以

x = \dfrac{1 - \cos\theta}{\sin\theta}

。

沿边

CD

，有

\begin{gathered} 1 = CF + FH + HD \\ = x + \sin\theta + x\cos\theta\text{。} \end{gathered}

代入

x

，得

\begin{gathered} 1 = \dfrac{(1-\cos\theta)(1+\cos\theta)}{\sin\theta} \\ {}+ \sin\theta \\ = \dfrac{\sin^2\theta}{\sin\theta} \\ {}+ \sin\theta \\ = 2\sin\theta\text{。} \end{gathered}

因此

\sin\theta = \tfrac12

，所以

\theta = 30^\circ

，并且

x = \dfrac{1 - \frac{\sqrt3}{2}}{\frac12} = 2 - \sqrt3\text{。}

所以正确答案是 C。

Let

x = BE = GH = CF

and

\theta = \angle DHG = \angle AGJ

= \angle FKH,

with

AD = GJ = HK = 1.

In right triangle

GDH,

x \sin\theta = DG = 1 - \cos\theta,

so

x = \dfrac{1 - \cos\theta}{\sin\theta}.

Along side

CD,

\begin{gathered} 1 = CF + FH + HD \\ = x + \sin\theta + x\cos\theta. \end{gathered}

Substituting for

x

gives

\begin{gathered} 1 = \dfrac{(1-\cos\theta)(1+\cos\theta)}{\sin\theta} \\ {}+ \sin\theta \\ = \dfrac{\sin^2\theta}{\sin\theta} \\ {}+ \sin\theta \\ = 2\sin\theta. \end{gathered}

Hence

\sin\theta = \tfrac12,

so

\theta = 30^\circ

and

x = \dfrac{1 - \frac{\sqrt3}{2}}{\frac12} = 2 - \sqrt3.

Thus, the correct answer is C.

其他年份的第 21 题