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Difference between revisions of "1976 AHSME Problems/Problem 27"
(The origin sol has flaws in the signs. It needs a reconstruction.)
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==Solution==
 
==Solution==
−
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
+
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> Clearly, <math>x</math> and <math>y</math> are both positive.
 +
 
 +
Note that
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
 
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} \\
Line 15: Line 17:
 
&=\frac{\left(\sqrt{5}+2\right)+2+\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>
 +
from which <math>x=\sqrt{2}.</math>
 +
 
 +
On the other hand, note that
 +
<cmath>\begin{align*}
 +
y&=\sqrt{3-2\sqrt{2}} \\
 +
&=\sqrt{2-2\sqrt{2}+1} \\
 +
&=\sqrt{\left(\sqrt{2}-1\right)^2} \\
 +
&=\sqrt{2}-1.
 
\end{align*}</cmath>
 
\end{align*}</cmath>
−
It is clear that <math>x>0,</math> from which <math>x=\sqrt{2}.</math>
+
Finally, the answer is <math>N=x-y=\boxed{\textbf{(A) }1}.</math>
  
−
~Someonenumber011 (Solution)
+
~Someonenumber011 (Fundamental Logic)
  
−
~MRENTHUSIASM (Revision)
+
~MRENTHUSIASM (Reconstruction)
  
 
== See also ==
 
== See also ==

Revision as of 01:36, 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 $y=\sqrt{3-2\sqrt{2}}.$ Clearly, $x$ and $y$ are both positive.

Note that \begin{align*} 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+\left(\sqrt{5}-2\right)}{\sqrt{5}+1} \\ &=\frac{2\sqrt{5}+2}{\sqrt{5}+1} \\ &=2, \end{align*} from which $x=\sqrt{2}.$

On the other hand, note that \begin{align*} y&=\sqrt{3-2\sqrt{2}} \\ &=\sqrt{2-2\sqrt{2}+1} \\ &=\sqrt{\left(\sqrt{2}-1\right)^2} \\ &=\sqrt{2}-1. \end{align*} Finally, the answer is $N=x-y=\boxed{\textbf{(A) }1}.$

~Someonenumber011 (Fundamental Logic)

~MRENTHUSIASM (Reconstruction)

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

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