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

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Note that
 
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>
 
<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>
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We find each term in the numerator separately:
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We rewrite each term in the numerator separately:
 
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<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 simplify:
 
   <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 simplify:

Revision as of 02:43, 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 $y=\sqrt{3-2\sqrt{2}}.$ 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)

Solution 2

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 \[x=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}\cdot\frac{\sqrt{\sqrt{5}-1}}{\sqrt{\sqrt{5}-1}}=\frac{\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}}{2}. \hspace{15mm} (\bigstar)\] 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 simplify: \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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