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>. | ||
− | 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}= | + | 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== |
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 is the ratio of the lesser part to the greater part, then the value of
is
Solution
Let be the length of the shorter segment and be the length of the longer segment. We're given that . Cross-multiplying, we find that . Now we divide both sides by to get . Therefore, .
From this, we have . Dividing both sides by , we get . Therefore, . Finally, we have
See Also
1974 AHSME (Problems • Answer Key • Resources) | ||
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