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

Revision as of 02:45, 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:

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 $\sqrt{3+\sqrt{5}}=\sqrt{\frac12}+\sqrt{\frac52}=\frac{\sqrt2}{2}+\frac{\sqrt{10}}{2}.$

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

@@ Line 41: / Line 41: @@
 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 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 rearrange:
 <cmath>\begin{align*}
 a+b+2\sqrt{ab}&=3+\sqrt{5} \\

1976 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 26	Followed by Problem 28
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Difference between revisions of "1976 AHSME Problems/Problem 27"

Revision as of 02:45, 8 September 2021

Contents

Problem

Solution 1

Solution 2

See also