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Difference between revisions of "1974 AHSME Problems/Problem 30"

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Let <math> w </math> be the length of the shorter segment and <math> l </math> be the length of the longer segment. We're given that <math> \frac{w}{l}=\frac{l}{w+l} </math>. Cross-multiplying, we find that <math> w^2+wl=l^2\implies w^2+wl-l^2=0 </math>. Now we divide both sides by <math> l^2 </math> to get <math> \left(\frac{w}{l}\right)^2+\left(\frac{w}{l}\right)-1=0 </math>. Therefore, <math> R^2+R-1=0 </math>.
 
Let <math> w </math> be the length of the shorter segment and <math> l </math> be the length of the longer segment. We're given that <math> \frac{w}{l}=\frac{l}{w+l} </math>. Cross-multiplying, we find that <math> w^2+wl=l^2\implies w^2+wl-l^2=0 </math>. Now we divide both sides by <math> l^2 </math> to get <math> \left(\frac{w}{l}\right)^2+\left(\frac{w}{l}\right)-1=0 </math>. Therefore, <math> R^2+R-1=0 </math>.
  
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From this, we have <math> R^2=-R+1 </math>. Dividing both sides by <math> R </math>, we get <math> R=-1+\frac{1}{R}\implies R^{-1}=R+1 </math>. Therefore, <math> R^2+R^{-1}=-R+1+R+1=2 </math>. Finally, we have <cmath> 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}}. </cmath>
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From this, we have <math> R^2=-R+1 </math>. Dividing both sides by <math> R </math>, we get <math> R=-1+\frac{1}{R}\implies R^{-1}=R+1 </math>. Therefore, <math> R^2+R^{-1}=-R+1+R+1=2 </math>. Finally, we have <cmath> R^{[R^{(R^2+R^{-1})}+R^{-1}]}+R^{-1}=R^{[R^2+R^{-1}]}+R^{-1}=R^2+R^{-1}= \boxed{\textbf{(A)}2}. </cmath>
  
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1974|num-b=29|after=Last Problem}}
 
{{AHSME box|year=1974|num-b=29|after=Last Problem}}
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 18:22, 19 March 2014

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 \[R^{[R^{(R^2+R^{-1})}+R^{-1}]}+R^{-1}=R^{[R^2+R^{-1}]}+R^{-1}=R^2+R^{-1}= \boxed{\textbf{(A)}2}.\]

See Also

1974 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 29 		Followed by
Last Problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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