1951 AMC 12 第 48 题

Question

内接于半圆的正方形面积与内接于整圆的正方形面积之比为：

The area of a square inscribed in a semicircle is to the area of the square inscribed in the entire circle as:

1:2

2:3

2:5

3:4

3:5

设圆的半径为

R

，内接于半圆的正方形边长为

s

Let the circle have radius

R

and the semicircle-square have side

s

对于内接于半圆的正方形，由一个上顶点可得

(\frac{s}{2})^2+s^2=R^2

For the semicircle-square, a top vertex gives

(\frac{s}{2})^2+s^2=R^2

对于内接于半圆的正方形，将其底边放在直径上。一个上顶点到圆心的水平距离为

\frac{s}{2}

，竖直距离为

s

，所以

\left(\frac s2\right)^2+s^2=R^2\text{，}

得到

s^2=\frac{4R^2}{5}

。内接于整圆的正方形对角线长为

2R

，故面积为

2R^2

。所求比为

\frac{\frac{4R^2}{5}}{2R^2}=\frac25\text{。}

因此，正确答案是 C。

For the square in the semicircle, put its base on the diameter. A top vertex has horizontal distance

\frac{s}{2}

from the center and vertical distance

s,

so

\left(\frac s2\right)^2+s^2=R^2,

giving

s^2=\frac{4R^2}{5}.

A square inscribed in the full circle has diagonal

2R,

hence area

2R^2.

The ratio is

\frac{\frac{4R^2}{5}}{2R^2}=\frac25.

Thus, the correct answer is C.

其他年份的第 48 题