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2016 AMC 12A 第 21 题

先试着解答 2016 AMC 12A 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 2016 AMC 12A 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.

一个四边形内接于半径为 2002200\sqrt{2} 的圆。该四边形的三条边长为 200200。第四条边长是多少？

A quadrilateral is inscribed in a circle of radius 2002.200\sqrt{2}. Three of the sides of this quadrilateral have length 200.200. What is the length of its fourth side?

200200

2002200\sqrt{2}

2003200\sqrt{3}

3002300\sqrt{2}

500500

答案：E
知识点：圆内接四边形余弦定理三角恒等式
难度评级：2040
小提示：

每条长 200200 的边对应圆心角 θ\theta，其中 cos⁡θ=34\cos\theta=\frac34。

Each side of length 200200 subtends a central angle θ\theta with cos⁡θ=34\cos\theta=\frac34

大提示：

第四条边对应 3θ3\theta；使用 cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos 3\theta=4\cos^3\theta-3\cos\theta。

The fourth chord has the same length as one subtending 3θ;3\theta; use cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos 3\theta=4\cos^3\theta-3\cos\theta

解答：

设长为 200200 的边所对的圆心角为 θ\theta，圆的半径为 R=2002R=200\sqrt2。对以圆心为顶点的等腰三角形应用余弦定理，得到 2002=2R2(1−cos⁡θ)=160000(1−cos⁡θ)， \begin{gathered} 200^2=2R^2(1-\cos\theta)\\ =160000(1-\cos\theta) \end{gathered}\text{，} 所以 cos⁡θ=34\cos\theta=\dfrac34。

三条相等的边占据三段圆心角均为 θ\theta 的连续弧，所以第四段弧的圆心角为 2π−3θ2\pi-3\theta。它所对的弦与圆心角为 3θ3\theta 的弦等长，并且 cos⁡3θ=4cos⁡3θ−3cos⁡θ=4⋅2764−94=−916。 \begin{gathered} \cos 3\theta=4\cos^3\theta-3\cos\theta\\ =4\cdot\dfrac{27}{64}-\dfrac94\\ =-\dfrac{9}{16} \end{gathered}\text{。} 它的长度平方为 2R2(1−cos⁡3θ)=160000(1+916)=160000⋅2516=250000， \begin{gathered} 2R^2(1-\cos 3\theta)\\ =160000\left(1+\dfrac{9}{16}\right)\\ =160000\cdot\dfrac{25}{16}\\ =250000 \end{gathered}\text{，} 所以第四条边长为 500500。

所以正确答案是 E。

Let θ\theta be the central angle subtending a side of length 200,200, with radius R=2002.R=200\sqrt2. By the law of cosines on the isosceles triangle from the center, 2002=2R2(1−cos⁡θ)=160000(1−cos⁡θ), \begin{gathered} 200^2=2R^2(1-\cos\theta)\\ =160000(1-\cos\theta), \end{gathered} so cos⁡θ=34.\cos\theta=\dfrac34.

The three equal sides use three consecutive arcs of angle θ,\theta, so the fourth arc has angle 2π−3θ.2\pi-3\theta. Its chord has the same length as a chord with central angle 3θ,3\theta, and cos⁡3θ=4cos⁡3θ−3cos⁡θ=4⋅2764−94=−916. \begin{gathered} \cos 3\theta=4\cos^3\theta-3\cos\theta\\ =4\cdot\dfrac{27}{64}-\dfrac94\\ =-\dfrac{9}{16}. \end{gathered} Its length squared is 2R2(1−cos⁡3θ)=160000(1+916)=160000⋅2516=250000, \begin{gathered} 2R^2(1-\cos 3\theta)\\ =160000\left(1+\dfrac{9}{16}\right)\\ =160000\cdot\dfrac{25}{16}\\ =250000, \end{gathered} so the fourth side is 500.500.

Thus, the correct answer is E.

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