1971 AMC 12 第 31 题

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31.

四边形 ABCDABCD 内接于一个圆,边 ADAD 是长为 44 的直径。若边 ABABBCBC 的长度都为 11,则边 CDCD 的长度为:

Quadrilateral ABCDABCD is inscribed in a circle with side AD,AD, a diameter of length 4.4. If sides ABAB and BCBC each have length 1,1, then side CDCD has length:

72\frac{7}{2}

522\frac{5\sqrt2}{2}

11\sqrt{11}

13\sqrt{13}

232\sqrt3

答案:A
知识点:三角恒等式
难度评级:2190
小提示:

等弦 ABABBCBC 所对的圆心角相等

Equal chords ABAB and BCBC subtend equal central angles

大提示:

若其中任一圆心角的一半为 tt,则 4sint=14\sin t=1,且 CD=4cos(2t)CD=4\cos(2t)

If half of either central angle is tt, then 4sint=14\sin t=1 and CD=4cos(2t)CD=4\cos(2t)

解答:

该圆的半径为 22。设等弦 ABABBCBC 所对的圆心角都为 2t2t。则 1=4sint 1=4\sin t\text{,} 所以 sint=14\sin t=\frac{1}{4}。沿半圆从 CCDD 的剩余圆心角为 π4t\pi-4t,故 CD=4sin(π4t2)=4cos(2t)=4(12116)=72 \begin{aligned} CD&=4\sin\left(\frac{\pi-4t}{2}\right)\\ &=4\cos(2t)\\ &=4\left(1-2\cdot\frac1{16}\right)\\ &=\frac72\text{。} \end{aligned}

因此,正确答案为 A

The circle has radius 2.2. Let the central angles subtending the equal chords ABAB and BCBC each be 2t.2t. Then 1=4sint, 1=4\sin t, so sint=14.\sin t=\frac{1}{4}. The remaining central angle from CC to DD along the semicircle is π4t,\pi-4t, hence CD=4sin(π4t2)=4cos(2t)=4(12116)=72. \begin{aligned} CD&=4\sin\left(\frac{\pi-4t}{2}\right)\\ &=4\cos(2t)\\ &=4\left(1-2\cdot\frac1{16}\right)\\ &=\frac72. \end{aligned}

Therefore, the correct answer is A.

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