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1974 AMC 12 Problem 30

Attempt Problem 30 of the 1974 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1974 AMC 12 solutions, or check the answer key.

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30.

A line segment is divided so that the lesser part is to the greater part as the greater part is to the whole. If RR is the ratio of the lesser part to the greater part, then the value of R[R(R2+R−1)+R−1]+R−1 R^{\left[R^{\left(R^2+R^{-1}\right)}+R^{-1}\right]}+R^{-1} is

22

2R2R

R−1R^{-1}

2+R−12+R^{-1}

2+R2+R

Answer: A
Concepts:quadraticexponentsubstitution
Difficulty rating: 2100
Small Hint:

Translate the division condition into R=11+RR=\frac{1}{1+R}

Big Hint:

From R2+R=1,R^2+R=1, derive R−1=R+1R^{-1}=R+1 and simplify from the innermost exponent outward

Solution:

Scale the greater part to 1,1, making the lesser part R.R. The condition gives R=11+R,R=\frac{1}{1+R}, so R2+R=1,R−1=R+1. R^2+R=1,\qquad R^{-1}=R+1. Consequently R2+R−1=R2+R+1=2. R^2+R^{-1}=R^2+R+1=2. Working outward, the exponent of the outer RR becomes R R2+R−1+R−1=R2+R−1=2. \begin{aligned} R^{\,R^2+R^{-1}}+R^{-1} &=R^2+R^{-1}\\ &=2. \end{aligned} The whole expression is therefore R2+R−1=2.R^2+R^{-1}=2.

Therefore, the correct answer is A.

← Problem 29#29
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