1974 AMC 12 第 30 题

Question

一条线段被分成两部分，使较短部分与较长部分之比等于较长部分与整条线段之比。若

R

是较短部分与较长部分之比，则

R^{\left[R^{\left(R^2+R^{-1}\right)}+R^{-1}\right]}+R^{-1}

的值为

A line segment is divided so that the lesser part is to the greater part as the greater part is to the whole. If

R

is the ratio of the lesser part to the greater part, then the value of

R^{\left[R^{\left(R^2+R^{-1}\right)}+R^{-1}\right]}+R^{-1}

is

2

2R

R^{-1}

2+R^{-1}

2+R

将分割条件写成

R=\frac{1}{1+R}

Translate the division condition into

R=\frac{1}{1+R}

由

R^2+R=1

推出

R^{-1}=R+1

，再从最内层指数向外化简

From

R^2+R=1,

derive

R^{-1}=R+1

and simplify from the innermost exponent outward

将较长部分缩放为

1

，则较短部分为

R

。由条件得

R=\frac{1}{1+R}

，所以

R^2+R=1,\qquad R^{-1}=R+1\text{。}

因此

R^2+R^{-1}=R^2+R+1=2\text{。}

从内向外化简，最外层

R

的指数变为

\begin{aligned} R^{\,R^2+R^{-1}}+R^{-1} &=R^2+R^{-1}\\ &=2 \end{aligned}\text{。}

所以整个表达式为

R^2+R^{-1}=2

。

因此，正确答案是 A。

Scale the greater part to

1,

making the lesser part

R.

The condition gives

R=\frac{1}{1+R},

so

R^2+R=1,\qquad R^{-1}=R+1.

Consequently

R^2+R^{-1}=R^2+R+1=2.

Working outward, the exponent of the outer

R

becomes

\begin{aligned} R^{\,R^2+R^{-1}}+R^{-1} &=R^2+R^{-1}\\ &=2. \end{aligned}

The whole expression is therefore

R^2+R^{-1}=2.

Therefore, the correct answer is A.

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