Difference between revisions of "1976 AHSME Problems/Problem 27"
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x^2&=\frac{\left(\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}\right)^2}{\left(\sqrt{\sqrt{5}+1}\right)^2} \\ | x^2&=\frac{\left(\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}\right)^2}{\left(\sqrt{\sqrt{5}+1}\right)^2} \\ | ||
&=\frac{\left(\sqrt{5}+2\right)+2\cdot\left(\sqrt{\sqrt{5}+2}\cdot\sqrt{\sqrt{5}-2}\right)+\left(\sqrt{5}-2\right)}{\sqrt{5}+1} \\ | &=\frac{\left(\sqrt{5}+2\right)+2\cdot\left(\sqrt{\sqrt{5}+2}\cdot\sqrt{\sqrt{5}-2}\right)+\left(\sqrt{5}-2\right)}{\sqrt{5}+1} \\ | ||
| + | &=\frac{\left(\sqrt{5}+2\right)+2+\left(\sqrt{5}-2\right)}{\sqrt{5}+1} \\ | ||
&=\frac{2\sqrt{5}+2}{\sqrt{5}+1} \\ | &=\frac{2\sqrt{5}+2}{\sqrt{5}+1} \\ | ||
&=2. | &=2. | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
| + | It is clear that <math>x>0,</math> from which <math>x=\sqrt{2}.</math> | ||
~Someonenumber011 (Solution) | ~Someonenumber011 (Solution) | ||
Revision as of 02:17, 8 September 2021
Problem
If ![\[N=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}},\]](http://latex.artofproblemsolving.com/f/0/8/f0884a5496b3ee2348e2a6edeacfa5a6879de8f5.png)
then 
equals
Solution
Let 
and 
Note that
It is clear that 
from which 
~Someonenumber011 (Solution)
~MRENTHUSIASM (Revision)
See also
| 1976 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 26 |
Followed by Problem 28 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
