1973 AMC 12 第 15 题

Question

从半径为

6

的圆中截取一个圆心角为锐角

\theta

的扇形。该扇形的外接圆半径为

A sector with acute central angle

\theta

is cut from a circle of radius

6.

The radius of the circle circumscribed about the sector is

3\cos\theta

3\sec\theta

3\cos\frac12\theta

3\sec\frac12\theta

3

将扇形的两条半径与弦看作一个等腰三角形

Treat the sector’s two radii and chord as an isosceles triangle

用

\frac{\theta}{2}

表示弦长，再应用正弦定理的扩展形式

Express the chord using

\frac{\theta}{2},

then apply the extended sine rule

两条半径与扇形的弦组成一个腰长为

6

、顶角为

\theta

的等腰三角形。其底边长为

12\sin(\frac{\theta}{2})

。若

R

为该三角形的外接圆半径，则扩展正弦定理给出

2R=\frac{12\sin(\frac{\theta}{2})}{\sin\theta}\text{。}

由于

\sin\theta=2\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})

，

R=3\sec\frac{\theta}{2}\text{。}

所以正确答案是 D。

The two radii and the sector’s chord form an isosceles triangle with equal sides

6

and vertex angle

\theta.

Its base has length

12\sin(\frac{\theta}{2}).

If

R

is the triangle’s circumradius, the extended sine rule gives

2R=\frac{12\sin(\frac{\theta}{2})}{\sin\theta}.

Since

\sin\theta=2\sin(\frac{\theta}{2})\cos(\frac{\theta}{2}),

R=3\sec\frac{\theta}{2}.

Therefore, the correct answer is D.

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