1967 AMC 12 第 3 题

Question

一个等边三角形的边长为

s

。该三角形内切一个圆，该圆又内接一个正方形。正方形的面积为：

The side of an equilateral triangle is

s.

A circle is inscribed in the triangle and a square is inscribed in the circle. The area of the square is:

\dfrac{s^2}{24}

\dfrac{s^2}{6}

\dfrac{s^2\sqrt2}{6}

\dfrac{s^2\sqrt3}{6}

\dfrac{s^2}{3}

边长为

s

的等边三角形，其内切圆半径为

\frac{s\sqrt3}{6}

The inradius of an equilateral triangle of side

s

is

\frac{s\sqrt3}{6}

正方形的对角线等于圆的直径

The square’s diagonal is the circle’s diameter

内切圆半径为

r=\frac{s\sqrt3}{6}

。内接正方形的对角线长为

2r

，所以其面积为该对角线平方的一半：

\begin{aligned} \frac{(2r)^2}{2} &=2r^2\\ &=2\left(\frac{s\sqrt3}{6}\right)^2\\ &=\frac{s^2}{6} \end{aligned}\text{。}

因此，正确答案是 B。

The inradius is

r=\frac{s\sqrt3}{6}.

The inscribed square has diagonal

2r,

so its area is half the square of that diagonal:

\begin{aligned} \frac{(2r)^2}{2} &=2r^2\\ &=2\left(\frac{s\sqrt3}{6}\right)^2\\ &=\frac{s^2}{6}. \end{aligned}

Therefore, the correct answer is B.

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