1976 AHSME Problems/Problem 27

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 $\begin{align*} x&=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}\cdot\frac{\sqrt{\sqrt{5}+1}}{\sqrt{\sqrt{5}+1}} \end{align*}$

1976 AHSME (Problems • Answer Key • Resources)
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1976 AHSME Problems/Problem 27

Contents

Problem

Solution 1

Solution 2

See also