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Difference between revisions of "1976 AHSME Problems/Problem 27"
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==Problem==
 
==Problem==
−
If <math>N=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}</math>, then <math>N</math> equals
+
If <cmath>N=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}},</cmath> then <math>N</math> equals
  
−
<math>\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}</math>
+
<math>\textbf{(A) }1\qquad
−
==Solution==
+
\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}   </math>  
  
−
We will split this problem into two parts: The fraction on the left and the square root on the right.
+
==Solution 1==
 +
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>
  
−
Starting with the fraction on the left, begin by squaring the numerator and putting a square root around it. It becomes
+
Note that
 +
<cmath>\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*}</cmath>
 +
Since <math>x>0,</math> we have <math>x=\sqrt{2}.</math>
  
−
<math>\frac{\sqrt{(\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2})^2}}{\sqrt{\sqrt{5}+1}} = \frac{\sqrt{(\sqrt{5}+2)+(\sqrt{5}-2)+2(
+
On the other hand, note that
−
\sqrt{(\sqrt{5}+2)(\sqrt{5}-2)}}}{\sqrt{\sqrt{5}+1}} = \frac{\sqrt{2\sqrt5+2\sqrt{(\sqrt5)^2-2^2}}}{\sqrt{\sqrt{5}+1}} =
+
<cmath>\begin{align*}
−
\frac{\sqrt{2\sqrt5+2\sqrt{1}}}{\sqrt{\sqrt5+1}} =
+
y^2&=3-2\sqrt{2} \\
−
\frac{\sqrt{2\sqrt5+2}}{\sqrt{\sqrt5+1}} = \frac{(\sqrt{2})(\sqrt{\sqrt5+1})}{\sqrt{\sqrt5+1}} = \sqrt2</math>.
+
&=2-2\sqrt{2}+1 \\
 +
&=\left(\sqrt{2}-1\right)^2.
 +
\end{align*}</cmath>
 +
Since <math>y>0,</math> we have <math>y=\sqrt{2}-1.</math>
  
−
Now for the right side.
+
Finally, the answer is <cmath>N=x-y=\boxed{\textbf{(A) }1}.</cmath>
  
−
<math>\sqrt{3-2\sqrt2} = \sqrt{(1-\sqrt2)^2} = 1-\sqrt2</math>
+
~Someonenumber011 (Fundamental Logic)
  
−
Putting it all together gives:
+
~MRENTHUSIASM (Reconstruction)
  
−
<math>(\sqrt2)+(1-\sqrt2)=\boxed{(A)  1}.</math>
+
==Solution 2==
 +
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>
  
−
~Someonenumber011
+
Note that
 +
<cmath>x=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}\cdot\frac{\sqrt{\sqrt{5}-1}}{\sqrt{\sqrt{5}-1}}=\frac{\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}}{2}. \hspace{15mm} (\bigstar)</cmath>
 +
We rewrite each term in the numerator separately:
 +
<ol style="margin-left: 1.5em;">
 +
  <li>Let <math>\sqrt{a}+\sqrt{b}=\sqrt{3+\sqrt{5}}</math> for some nonnegative rational numbers <math>a</math> and <math>b.</math> We square both sides of this equation, then rearrange:
 +
<cmath>\begin{align*}
 +
a+b+2\sqrt{ab}&=3+\sqrt{5} \\
 +
a+b+\sqrt{4ab}&=3+\sqrt{5}.
 +
\end{align*}</cmath>
 +
It follows that
 +
<cmath>\begin{align*}
 +
a+b&=3, \\
 +
ab&=\frac54.
 +
\end{align*}</cmath>
 +
By inspection, we get <math>\{a,b\}=\left\{\frac12,\frac52\right\}.</math> Alternatively, we conclude that <math>a</math> and <math>b</math> are the solutions to the quadratic equation <math>t^2-3t+\frac54=0</math> by Vieta's Formulas, in which <math>t=\frac12,\frac52.</math> <p> Therefore, we obtain <cmath>\sqrt{3+\sqrt{5}}=\sqrt{\frac12}+\sqrt{\frac52}=\frac{\sqrt2}{2}+\frac{\sqrt{10}}{2}.</cmath></li><p>
 +
  <li>Similarly, we obtain <cmath>\sqrt{7-3\sqrt{5}}=\frac{3\sqrt2}{2}-\frac{\sqrt{10}}{2}.</cmath></li><p>
 +
</ol>
 +
Substituting these results into <math>(\bigstar),</math> we have <cmath>x=\frac{\left(\frac{\sqrt2}{2}+\frac{\sqrt{10}}{2}\right)+\left(\frac{3\sqrt2}{2}-\frac{\sqrt{10}}{2}\right)}{2}=\sqrt2.</cmath>
 +
On the other hand, we have <cmath>y=\sqrt2-1</cmath> by the argument of either Solution 1 or Solution 2.
 +
 
 +
Finally, the answer is <cmath>N=x-y=\boxed{\textbf{(A) }1}.</cmath>
 +
 
 +
~MRENTHUSIASM
 +
 
 +
== See also ==
 +
 
 +
{{AHSME box|year=1976|num-b=26|num-a=28}}
 +
{{MAA Notice}}
 +
[[Category:AHSME]]
 +
[[Category:AHSME Problems]]

Latest revision as of 12:41, 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 1

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{\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*} Since $x>0,$ we have $x=\sqrt{2}.$

On the other hand, note that \begin{align*} y^2&=3-2\sqrt{2} \\ &=2-2\sqrt{2}+1 \\ &=\left(\sqrt{2}-1\right)^2. \end{align*} Since $y>0,$ we have $y=\sqrt{2}-1.$

Finally, the answer is \[N=x-y=\boxed{\textbf{(A) }1}.\]

~Someonenumber011 (Fundamental Logic)

~MRENTHUSIASM (Reconstruction)

Solution 2

Let $x=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}$ and $y=\sqrt{3-2\sqrt{2}}.$

Note that \[x=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}\cdot\frac{\sqrt{\sqrt{5}-1}}{\sqrt{\sqrt{5}-1}}=\frac{\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}}{2}. \hspace{15mm} (\bigstar)\] We rewrite each term in the numerator separately:

  1. Let $\sqrt{a}+\sqrt{b}=\sqrt{3+\sqrt{5}}$ for some nonnegative rational numbers $a$ and $b.$ We square both sides of this equation, then rearrange: \begin{align*} a+b+2\sqrt{ab}&=3+\sqrt{5} \\ a+b+\sqrt{4ab}&=3+\sqrt{5}. \end{align*} It follows that \begin{align*} a+b&=3, \\ ab&=\frac54. \end{align*} By inspection, we get $\{a,b\}=\left\{\frac12,\frac52\right\}.$ Alternatively, we conclude that $a$ and $b$ are the solutions to the quadratic equation $t^2-3t+\frac54=0$ by Vieta's Formulas, in which $t=\frac12,\frac52.$

    Therefore, we obtain \[\sqrt{3+\sqrt{5}}=\sqrt{\frac12}+\sqrt{\frac52}=\frac{\sqrt2}{2}+\frac{\sqrt{10}}{2}.\]

  2. Similarly, we obtain \[\sqrt{7-3\sqrt{5}}=\frac{3\sqrt2}{2}-\frac{\sqrt{10}}{2}.\]

Substituting these results into $(\bigstar),$ we have \[x=\frac{\left(\frac{\sqrt2}{2}+\frac{\sqrt{10}}{2}\right)+\left(\frac{3\sqrt2}{2}-\frac{\sqrt{10}}{2}\right)}{2}=\sqrt2.\] On the other hand, we have \[y=\sqrt2-1\] by the argument of either Solution 1 or Solution 2.

Finally, the answer is \[N=x-y=\boxed{\textbf{(A) }1}.\]

~MRENTHUSIASM

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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