Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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 $y=\sqrt{3-2\sqrt{2}}.$

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1976_AHSME_Problems/Problem_27&oldid=161875"