Question

一个四边形内接于半径为 $200\sqrt{2}$ 的圆。该四边形的三条边长为 $200$ 。第四条边长是多少？

A quadrilateral is inscribed in a circle of radius $200\sqrt{2}.$ Three of the sides of this quadrilateral have length $200.$ What is the length of its fourth side?

$200$

$200\sqrt{2}$

$200\sqrt{3}$

$300\sqrt{2}$

$500$

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

解答：

设长 $200$ 的边所对的圆心角为 $\theta$ ，圆的半径为 $R=200\sqrt2$ 。在圆心构成的等腰三角形中应用余弦定理，得到所以 $\cos\theta=\dfrac34$ 。 $\begin{gathered} 200^2=2R^2(1-\cos\theta)\\ =160000(1-\cos\theta), \end{gathered}$

第四条边所对的圆心角为 $3\theta$ ，并且它的边长平方为所以第四条边长为 $500$ 。 $\theta$ ， $2\pi-3\theta$ 。 $\begin{gathered} \cos 3\theta=4\cos^3\theta-3\cos\theta\\ =4\cdot\dfrac{27}{64}-\dfrac94\\ =-\dfrac{9}{16}. \end{gathered}$ $\begin{gathered} 2R^2(1-\cos 3\theta)\\ =160000\left(1+\dfrac{9}{16}\right)\\ =160000\cdot\dfrac{25}{16}\\ =250000, \end{gathered}$

所以正确答案是 E。

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

The three equal sides use three consecutive arcs of angle $\theta,$ so the fourth arc has angle $2\pi-3\theta.$ Its chord has the same length as a chord with central angle $3\theta,$ and $\begin{gathered} \cos 3\theta=4\cos^3\theta-3\cos\theta\\ =4\cdot\dfrac{27}{64}-\dfrac94\\ =-\dfrac{9}{16}. \end{gathered}$ Its length squared is $\begin{gathered} 2R^2(1-\cos 3\theta)\\ =160000\left(1+\dfrac{9}{16}\right)\\ =160000\cdot\dfrac{25}{16}\\ =250000, \end{gathered}$ so the fourth side is $500.$

Thus, the correct answer is E.

2016 AMC 12A 第 21 题

答案：E

难度评级：2040

解答：

其他年份的第 21 题