1974 AHSME Problems/Problem 30

Problem

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

$R^{[R^{(R^2+R^{-1})}+R^{-1}]}+R^{-1}$

is

$\mathrm{(A)\ } 2 \qquad \mathrm{(B) \ }2R \qquad \mathrm{(C) \ } R^{-1} \qquad \mathrm{(D) \ } 2+R^{-1} \qquad \mathrm{(E) \ }2+R$

Solution

Let $w$ be the length of the shorter segment and $l$ be the length of the longer segment. We're given that $\frac{w}{l}=\frac{l}{w+l}$ . Cross-multiplying, we find that $w^2+wl=l^2\implies w^2+wl-l^2=0$ . Now we divide both sides by $l^2$ to get $\left(\frac{w}{l}\right)^2+\left(\frac{w}{l}\right)-1=0$ . Therefore, $R^2+R-1=0$ .

From this, we have $R^2=-R+1$ . Dividing both sides by $R$ , we get $R=-1+\frac{1}{R}\implies R^{-1}=R+1$ . Therefore, $R^2+R^{-1}=-R+1+R+1=2$ . Finally, we have $R^{[R^{(R^2+R^{-1})}+R^{-1}]}+R^{-1}=R^{[R^2+R^{-1}]}+R^{-1}=R^2+R^{-1}=2, \boxed{\text{A}}.$

1974 AHSME (Problems • Answer Key • Resources)
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1974 AHSME Problems/Problem 30

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