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2014 AMC 12B 第 21 题

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所有题目均经美国数学协会（MAA）官方合法授权使用。

21.

图中，ABCDABCD 是边长为 11 的正方形。矩形 JKHGJKHG 和 EBCFEBCF 全等。BEBE 是多少？

In the figure, ABCDABCD is a square of side length 1.1. The rectangles JKHGJKHG and EBCFEBCF are congruent. What is BE?BE?

12(6−2)\dfrac12(\sqrt{6} - 2)

14\dfrac14

2−32 - \sqrt{3}

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

1−221 - \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
解答：

设 x=BE=GH=CFx = BE = GH = CF，并设 θ=∠DHG=∠AGJ\theta = \angle DHG = \angle AGJ =∠FKH= \angle FKH， 其中 AD=GJ=HK=1AD = GJ = HK = 1。 在直角三角形 GDHGDH 中，xsin⁡θ=DG=1−cos⁡θx \sin\theta = DG = 1 - \cos\theta， 所以 x=1−cos⁡θsin⁡θx = \dfrac{1 - \cos\theta}{\sin\theta}。

沿边 CDCD，有 代入 xx，得 1=CF+FH+HD=x+sin⁡θ+xcos⁡θ. \begin{gathered} 1 = CF + FH + HD \\ = x + \sin\theta + x\cos\theta. \end{gathered} 1=(1−cos⁡θ)(1+cos⁡θ)sin⁡θ+sin⁡θ=sin⁡2θsin⁡θ+sin⁡θ=2sin⁡θ. \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}

因此 sin⁡θ=12\sin\theta = \tfrac12， 所以 θ=30∘\theta = 30^\circ，并且 x=1−3212=2−3. x = \dfrac{1 - \frac{\sqrt3}{2}}{\frac12} = 2 - \sqrt3.

所以正确答案是 C。

Let x=BE=GH=CFx = BE = GH = CF and θ=∠DHG=∠AGJ\theta = \angle DHG = \angle AGJ =∠FKH,= \angle FKH, with AD=GJ=HK=1.AD = GJ = HK = 1. In right triangle GDH,GDH, xsin⁡θ=DG=1−cos⁡θ,x \sin\theta = DG = 1 - \cos\theta, so x=1−cos⁡θsin⁡θ.x = \dfrac{1 - \cos\theta}{\sin\theta}.

Along side CD,CD, 1=CF+FH+HD=x+sin⁡θ+xcos⁡θ. \begin{gathered} 1 = CF + FH + HD \\ = x + \sin\theta + x\cos\theta. \end{gathered} Substituting for xx gives 1=(1−cos⁡θ)(1+cos⁡θ)sin⁡θ+sin⁡θ=sin⁡2θsin⁡θ+sin⁡θ=2sin⁡θ. \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⁡θ=12,\sin\theta = \tfrac12, so θ=30∘\theta = 30^\circ and x=1−3212=2−3. x = \dfrac{1 - \frac{\sqrt3}{2}}{\frac12} = 2 - \sqrt3.

Thus, the correct answer is C.

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