1959 AMC 12 第 21 题

Question

若内接于一个圆的等边三角形周长为

p

，则该圆的面积为：

If

p

is the perimeter of an equilateral triangle inscribed in a circle, the area of the circle is:

\dfrac{\pi p^2}{3}

\dfrac{\pi p^2}{9}

\dfrac{\pi p^2}{27}

\dfrac{\pi p^2}{81}

\dfrac{\pi p^2\sqrt3}{27}

三角形边长为

\frac{p}{3}

The triangle side length is

\frac{p}{3}

边长为

s

的等边三角形外接圆半径为

\frac{s}{\sqrt3}

An equilateral triangle with side

s

has circumradius

\frac{s}{\sqrt3}

边长为

\frac{p}{3}

，所以外接圆半径为

R=\frac{\frac{p}{3}}{\sqrt3}=\frac{p}{3\sqrt3}\text{。}

因此圆的面积为

\pi R^2=\frac{\pi p^2}{27}\text{。}

因此，正确答案是 C。

The side length is

\frac{p}{3},

so the circumradius is

R=\frac{\frac{p}{3}}{\sqrt3}=\frac{p}{3\sqrt3}.

Therefore the circle’s area is

\pi R^2=\frac{\pi p^2}{27}.

Thus, the correct answer is C.

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