Difference between revisions of "1976 AHSME Problems/Problem 27"

(Centered the important equation: https://files.eric.ed.gov/fulltext/ED239856.pdf Also, I am thinking of making the solution clearer using algebra.)
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==Solution==
 
==Solution==
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Let <math>x=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}</math> and <math>y=\sqrt{3-2\sqrt{2}}.</math> Note that
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<cmath>\begin{align*}
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x^2&=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}
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\end{align*}</cmath>
  
 
~Someonenumber011 (Solution)
 
~Someonenumber011 (Solution)
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{{AHSME box|year=1976|num-b=26|num-a=28}}
 
{{AHSME box|year=1976|num-b=26|num-a=28}}
 
{{MAA Notice}}
 
{{MAA Notice}}
[[Category:AHSME]][[Category:AHSME Problems]]
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[[Category:AHSME]]
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[[Category:AHSME Problems]]

Revision as of 01:52, 8 September 2021

Problem

If \[N=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}},\] then $N$ equals

$\textbf{(A) }1\qquad \textbf{(B) }2\sqrt{2}-1\qquad \textbf{(C) }\frac{\sqrt{5}}{2}\qquad \textbf{(D) }\sqrt{\frac{5}{2}}\qquad \textbf{(E) }\text{none of these}$

Solution

Let $x=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}$ and $y=\sqrt{3-2\sqrt{2}}.$ Note that \begin{align*} x^2&=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}} \end{align*}

~Someonenumber011 (Solution)

~MRENTHUSIASM (Revision)

See also

1976 AHSME (ProblemsAnswer KeyResources)
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

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