1966 AMC 12 第 4 题

Question

圆

\mathrm{I}

外接于一个给定的正方形，圆

\mathrm{II}

内切于该正方形。若

r

是圆

\mathrm{I}

与圆

\mathrm{II}

的面积之比，则

r

等于：

Circle

\mathrm{I}

is circumscribed about a given square and circle

\mathrm{II}

is inscribed in the given square. If

r

is the ratio of the area of circle

\mathrm{I}

to that of circle

\mathrm{II},

then

r

equals:

\sqrt2

2

\sqrt3

2\sqrt2

2\sqrt3

设正方形边长为

s

，比较两个半径

\frac{s}{\sqrt2}

与

\frac{s}{2}

For square side

s,

compare radii

\frac{s}{\sqrt2}

and

\frac{s}{2}

圆的面积与半径的平方成正比

Circle areas are proportional to the squares of their radii

设正方形边长为

s

，则外接圆半径为

\frac{s}{\sqrt2}

，内切圆半径为

\frac{s}{2}

。因此

r=\frac{\pi(\frac{s}{\sqrt2})^2}{\pi(\frac{s}{2})^2}=2\text{。}

因此，正确答案是 B。

For square side

s,

the outer circle has radius

\frac{s}{\sqrt2},

while the inner circle has radius

\frac{s}{2}.

Hence

r=\frac{\pi(\frac{s}{\sqrt2})^2}{\pi(\frac{s}{2})^2}=2.

Therefore, the correct answer is B.

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