1989 AMC 12 第 19 题

Question

一个三角形内接于圆。三角形的三个顶点将圆周分成弧长分别为

3

、

4

和

5

的三段弧。这个三角形的面积是多少？

A triangle is inscribed in a circle. The vertices of the triangle divide the circle into three arcs of lengths

3,

4,

and

5.

What is the area of the triangle?

6

\frac{18}{\pi^2}

\frac9{\pi^2}(\sqrt3-1)

\frac9{\pi^2}(\sqrt3+1)

\frac9{\pi^2}(\sqrt3+3)

圆周长为

12

，先求出半径和三个圆心角

The circumference is

12,

so first find the radius and the three central angles

连接圆心与三个顶点，将该三角形分成三个以圆心为公共顶点的三角形

Split the triangle into three triangles having the circle’s center as a common vertex

半径为

R=\frac{12}{2\pi}=\frac{6}{\pi}

。弧长

3,4,5

对应的圆心角为

\frac{\pi}{2},\frac{2\pi}{3},\frac{5\pi}{6}

。这些角的正弦之和为

1+\frac{\sqrt3}{2}+\frac{1}{2}=\frac{3+\sqrt3}{2}

。从圆心将三角形分割后，其面积为

\begin{aligned} K&=\frac12\cdot\frac{36}{\pi^2} \cdot\frac{3+\sqrt3}{2}\\ &=\frac9{\pi^2}(\sqrt3+3) \end{aligned}\text{。}

所以正确答案是 E。

The radius is

R=\frac{12}{2\pi}=\frac{6}{\pi}.

The arc lengths

3,4,5

give central angles

\frac{\pi}{2},\frac{2\pi}{3},\frac{5\pi}{6}.

Their sines sum to

1+\frac{\sqrt3}{2}+\frac{1}{2}=\frac{3+\sqrt3}{2}.

Splitting the triangle at the center, its area is

\begin{aligned} K&=\frac12\cdot\frac{36}{\pi^2} \cdot\frac{3+\sqrt3}{2}\\ &=\frac9{\pi^2}(\sqrt3+3). \end{aligned}

Thus the correct answer is E.

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