1974 AMC 12 第 16 题

Question

半径为

r

的圆内切于一个等腰直角三角形，半径为

R

的圆外接于该三角形。则

\frac{R}{r}

等于

A circle of radius

r

is inscribed in a right isosceles triangle, and a circle of radius

R

is circumscribed about the triangle. Then

\frac{R}{r}

equals

1+\sqrt2

\frac{2+\sqrt2}{2}

\frac{\sqrt2-1}{2}

\frac{1+\sqrt2}{2}

2(2-\sqrt2)

设每条直角边长为

a

，用

a

表示两个半径

Let each leg have length

a

and express both radii in terms of

a

对直角三角形，外接圆半径是斜边的一半；内切圆半径为

\frac{a+a-a\sqrt2}{2}

For a right triangle, the circumradius is half the hypotenuse; the inradius is

\frac{a+a-a\sqrt2}{2}

两条直角边长均为

a

，斜边长为

a\sqrt2

，所以

\begin{aligned} R&=\frac{a\sqrt2}{2},\\ r&=\frac{a+a-a\sqrt2}{2} \end{aligned}\text{。}

因此

\begin{aligned} \frac Rr&=\frac{\sqrt2}{2-\sqrt2}\\ &=\frac1{\sqrt2-1}=1+\sqrt2 \end{aligned}\text{。}

因此，正确答案是 A。

With both legs of length

a

and hypotenuse

a\sqrt2,

\begin{aligned} R&=\frac{a\sqrt2}{2},\\ r&=\frac{a+a-a\sqrt2}{2}. \end{aligned}

Therefore

\begin{aligned} \frac Rr&=\frac{\sqrt2}{2-\sqrt2}\\ &=\frac1{\sqrt2-1}=1+\sqrt2. \end{aligned}

Therefore, the correct answer is A.

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