AMC12 2014 B

AMC12 2014 B · Q21

AMC12 2014 B · Q21. It mainly tests Circle theorems, Trigonometry (basic).

In the figure, $ABCD$ is a square of side length 1. The rectangles $JKHG$ and $EBCF$ are congruent. What is $BE$?

在图中，$ABCD$是一个边长为1的正方形。矩形$JKHG$和$EBCF$全等。$BE$的长度是多少？

(A) $\frac{1}{2}(\sqrt{6}-2)$ $\frac{1}{2}(\sqrt{6}-2)$

(B) $\frac{1}{4}$ $\frac{1}{4}$

(D) $\frac{\sqrt{3}}{6}$ $\frac{\sqrt{3}}{6}$

(E) $1 - \frac{\sqrt{2}}{2}$ $1 - \frac{\sqrt{2}}{2}$

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Let $x = BE = GH = CF$, and let $\theta = \angle DHG = \angle AGJ = \angle FKH$. Note that $AD = GJ = HK = 1$. In right triangle $GDH$, $x\sin\theta = DG = 1 - AG = 1 - \cos\theta$, so $x = \dfrac{1-\cos\theta}{\sin\theta}$. Then $1 = CD = CF + FH + HD = x + \sin\theta + x\cos\theta$. Substituting for $x$ gives $$ 1=\frac{1-\cos\theta}{\sin\theta}+\sin\theta+\frac{1-\cos\theta}{\sin\theta}\cdot\cos\theta $$ $$ =\frac{(1-\cos\theta)(1+\cos\theta)}{\sin\theta}+\sin\theta $$ $$ =\frac{\sin^2\theta}{\sin\theta}+\sin\theta=2\sin\theta. $$ It follows that $\sin\theta=\dfrac12$, so $\theta=30^\circ$, and $$ x=\frac{1-\frac{\sqrt3}{2}}{\frac12}=2-\sqrt3. $$

答案（C）：令 $x = BE = GH = CF$，并令 $\theta = \angle DHG = \angle AGJ = \angle FKH$。注意 $AD = GJ = HK = 1$。在直角三角形 $GDH$ 中，$x\sin\theta = DG = 1-AG = 1-\cos\theta$，因此 $x=\dfrac{1-\cos\theta}{\sin\theta}$。又有 $1 = CD = CF + FH + HD = x + \sin\theta + x\cos\theta$。代入 $x$ 得 $$ 1=\frac{1-\cos\theta}{\sin\theta}+\sin\theta+\frac{1-\cos\theta}{\sin\theta}\cdot\cos\theta $$ $$ =\frac{(1-\cos\theta)(1+\cos\theta)}{\sin\theta}+\sin\theta $$ $$ =\frac{\sin^2\theta}{\sin\theta}+\sin\theta=2\sin\theta. $$ 因此 $\sin\theta=\dfrac12$，所以 $\theta=30^\circ$，并且 $$ x=\frac{1-\frac{\sqrt3}{2}}{\frac12}=2-\sqrt3. $$

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