Difference between revisions of "1976 AHSME Problems/Problem 27"
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==Solution 1== | ==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> | + | 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> |
Note that | Note that | ||
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&=\frac{\left(\sqrt{5}+2\right)+2+\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} \\ | &=\frac{2\sqrt{5}+2}{\sqrt{5}+1} \\ | ||
| − | &=2 | + | &=2. |
\end{align*}</cmath> | \end{align*}</cmath> | ||
| − | + | Since <math>x>0,</math> we have <math>x=\sqrt{2}.</math> | |
On the other hand, note that | On the other hand, note that | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
| − | y&= | + | y^2&=3-2\sqrt{2} \\ |
| − | &= | + | &=2-2\sqrt{2}+1 \\ |
| − | &= | + | &=\left(\sqrt{2}-1\right)^2. |
| − | |||
\end{align*}</cmath> | \end{align*}</cmath> | ||
| + | Since <math>y>0,</math> we have <math>y=\sqrt{2}-1.</math> | ||
| + | |||
Finally, the answer is <math>N=x-y=\boxed{\textbf{(A) }1}.</math> | Finally, the answer is <math>N=x-y=\boxed{\textbf{(A) }1}.</math> | ||
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==Solution 2== | ==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> | + | 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> |
Note that | Note that | ||
Revision as of 12:28, 8 September 2021
Contents
Problem
If ![\[N=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}},\]](http://latex.artofproblemsolving.com/f/0/8/f0884a5496b3ee2348e2a6edeacfa5a6879de8f5.png)
then 
equals
Solution 1
Let 
and 
Note that
Since 
we have 
On the other hand, note that
Since 
we have 
Finally, the answer is 
~Someonenumber011 (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)
Solution 2
Let and
Note that
![\[x=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}\cdot\frac{\sqrt{\sqrt{5}-1}}{\sqrt{\sqrt{5}-1}}=\frac{\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}}{2}. \hspace{15mm} (\bigstar)\]](http://latex.artofproblemsolving.com/3/6/d/36d8ff4f73a033849cef445123a88dc57001a661.png)
We rewrite each term in the numerator separately:
- Let
for some nonnegative rational numbers 
and 
We square both sides of this equation, then rearrange:
It follows that
By inspection, we get 
Alternatively, we conclude that and 
are the solutions to the quadratic equation 
by Vieta's Formulas, in which 
Therefore, we obtain
![\[\sqrt{3+\sqrt{5}}=\sqrt{\frac12}+\sqrt{\frac52}=\frac{\sqrt2}{2}+\frac{\sqrt{10}}{2}.\]](//latex.artofproblemsolving.com/c/e/e/ceec8b0def0605f4ac4234786bdc25b9c1128f36.png)
- Similarly, we obtain
![\[\sqrt{7-3\sqrt{5}}=\frac{3\sqrt2}{2}-\frac{\sqrt{10}}{2}.\]](//latex.artofproblemsolving.com/2/8/7/287a2623f3c3b52a1e471d3120945427b8294faf.png)
for some nonnegative rational numbers 
and 
We square both sides of this equation, then rearrange:
It follows that
By inspection, we get 
Alternatively, we conclude that 
are the solutions to the quadratic equation 
by Vieta's Formulas, in which 
![\[\sqrt{3+\sqrt{5}}=\sqrt{\frac12}+\sqrt{\frac52}=\frac{\sqrt2}{2}+\frac{\sqrt{10}}{2}.\]](http://latex.artofproblemsolving.com/c/e/e/ceec8b0def0605f4ac4234786bdc25b9c1128f36.png)
![\[\sqrt{7-3\sqrt{5}}=\frac{3\sqrt2}{2}-\frac{\sqrt{10}}{2}.\]](http://latex.artofproblemsolving.com/2/8/7/287a2623f3c3b52a1e471d3120945427b8294faf.png)
Substituting these results into 
we have ![\[x=\frac{\left(\frac{\sqrt2}{2}+\frac{\sqrt{10}}{2}\right)+\left(\frac{3\sqrt2}{2}-\frac{\sqrt{10}}{2}\right)}{2}=\sqrt2.\]](http://latex.artofproblemsolving.com/2/c/7/2c7451af2648654ed70f528f1e4291e5b3b76235.png)
On the other hand, we have ![\[y=\sqrt2-1\]](http://latex.artofproblemsolving.com/d/2/9/d29de6bb314680d47f6334601c17134c6ab48cad.png)
by the argument of either Solution 1 or Solution 2.
Finally, the answer is
~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. 
